Interpreting Force Diagrams - AP Physics C: Electricity and Magnetism
Card 1 of 44
A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?
A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?
Tap to reveal answer
When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.
In addition, the rocket experiences the downward force of its own weight. This is given by:
We know that 
.
We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.
We know that 
(since 
is directed upwards and 
is directed downwards).
Finally we solve for acceleration:
When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.
In addition, the rocket experiences the downward force of its own weight. This is given by:
We know that
We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.
We know that
Finally we solve for acceleration:
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Three boxes tied by two ropes move across a frictionless surface pulled by a force 
as shown in the diagram. Which of the following is an expression of the acceleration of the system?
Three boxes tied by two ropes move across a frictionless surface pulled by a force
Tap to reveal answer
Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.
A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between 
and 
as 
. Similarly, we will label the tension experienced by the rope between 
and 
as 
.
For the box on the right (of mass 
) we know that the force 
pulls to the right and the rope pulls it to the left with a tension of 
.
We get the equation: 
. 
is the mass of the box and 
is its acceleration (which is the same for the other boxes).
For the middle box (of mass 
) we have a rope pulling on it on the right with tension 
and a rope pulling on the left with tension 
.
So we get the equation: 
. 
is the mass of the box and 
is its acceleration (which is the same for the other boxes).
Finally, for the box on the left (of mass 
) we have only one rope pulling it to the right with a tension 
.
So we get the equation: 
. 
is the mass of the box and 
is its acceleration (which is the same for the other boxes).
Now it is just a matter of simple substitution. We have three equations:
From them we can get an expression for the force. Isolate in the second equation.
Substitute the expression of from the first equation and simplify.
Use this value of in the third equation to get an expression for the force.
So we have:
Solve for acceleration.
Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.
A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between
For the box on the right (of mass
We get the equation:
For the middle box (of mass
So we get the equation:
Finally, for the box on the left (of mass
So we get the equation:
Now it is just a matter of simple substitution. We have three equations:
From them we can get an expression for the force. Isolate
Substitute the expression of
Use this value of
So we have:
Solve for acceleration.
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A box is being pushed against a wall by a force 
as shown in the picture. If the coefficient of static friction between the box and the wall is 
, which of the following expressions represents what the force must be for the box not to fall?
A box is being pushed against a wall by a force
Tap to reveal answer
This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force 
. The box experiences the downward force of its own weight given by 
. Finally, since the box is trying to slide down, friction 
opposes this motion. The following diagram shows these forces:
The key is that the box should NOT move.
For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.
Now, on the vertical axis we have fiction and the object's weight:
Finally, we use the equation for friction force to solve the problem.
Substitute the weight, since it is equal to the force of friction.
Isolate the normal force.
Since the normal force is equal to the applied force, this is our final expression.
This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force
The key is that the box should NOT move.
For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.
Now, on the vertical axis we have fiction and the object's weight:
Finally, we use the equation for friction force to solve the problem.
Substitute the weight, since it is equal to the force of friction.
Isolate the normal force.
Since the normal force is equal to the applied force, this is our final expression.
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A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?
A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?
Tap to reveal answer
You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.
We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.
We can solve for the horizontal component of the force using the applied force as the hypotenuse:
You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.
We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.
We can solve for the horizontal component of the force using the applied force as the hypotenuse:
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A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?
A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?
Tap to reveal answer
When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.
In addition, the rocket experiences the downward force of its own weight. This is given by:
We know that .
We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.
We know that (since is directed upwards and is directed downwards).
Finally we solve for acceleration:
When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.
In addition, the rocket experiences the downward force of its own weight. This is given by:
We know that
We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.
We know that
Finally we solve for acceleration:
← Didn't Know|Knew It →
Three boxes tied by two ropes move across a frictionless surface pulled by a force as shown in the diagram. Which of the following is an expression of the acceleration of the system?
Three boxes tied by two ropes move across a frictionless surface pulled by a force
Tap to reveal answer
Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.
A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between and as . Similarly, we will label the tension experienced by the rope between and as .
For the box on the right (of mass ) we know that the force pulls to the right and the rope pulls it to the left with a tension of .
We get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).
For the middle box (of mass ) we have a rope pulling on it on the right with tension and a rope pulling on the left with tension .
So we get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).
Finally, for the box on the left (of mass ) we have only one rope pulling it to the right with a tension .
So we get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).
Now it is just a matter of simple substitution. We have three equations:
From them we can get an expression for the force. Isolate in the second equation.
Substitute the expression of from the first equation and simplify.
Use this value of in the third equation to get an expression for the force.
So we have:
Solve for acceleration.
Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.
A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between
For the box on the right (of mass
We get the equation:
For the middle box (of mass
So we get the equation:
Finally, for the box on the left (of mass
So we get the equation:
Now it is just a matter of simple substitution. We have three equations:
From them we can get an expression for the force. Isolate
Substitute the expression of
Use this value of
So we have:
Solve for acceleration.
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A box is being pushed against a wall by a force as shown in the picture. If the coefficient of static friction between the box and the wall is , which of the following expressions represents what the force must be for the box not to fall?
A box is being pushed against a wall by a force
Tap to reveal answer
This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force . The box experiences the downward force of its own weight given by . Finally, since the box is trying to slide down, friction opposes this motion. The following diagram shows these forces:
The key is that the box should NOT move.
For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.
Now, on the vertical axis we have fiction and the object's weight:
Finally, we use the equation for friction force to solve the problem.
Substitute the weight, since it is equal to the force of friction.
Isolate the normal force.
Since the normal force is equal to the applied force, this is our final expression.
This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force
The key is that the box should NOT move.
For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.
Now, on the vertical axis we have fiction and the object's weight:
Finally, we use the equation for friction force to solve the problem.
Substitute the weight, since it is equal to the force of friction.
Isolate the normal force.
Since the normal force is equal to the applied force, this is our final expression.
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A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?
A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?
Tap to reveal answer
You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.
We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.
We can solve for the horizontal component of the force using the applied force as the hypotenuse:
You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.
We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.
We can solve for the horizontal component of the force using the applied force as the hypotenuse:
← Didn't Know|Knew It →
A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?
A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?
Tap to reveal answer
When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.
In addition, the rocket experiences the downward force of its own weight. This is given by:
We know that .
We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.
We know that (since is directed upwards and is directed downwards).
Finally we solve for acceleration:
When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.
In addition, the rocket experiences the downward force of its own weight. This is given by:
We know that
We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.
We know that
Finally we solve for acceleration:
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Three boxes tied by two ropes move across a frictionless surface pulled by a force as shown in the diagram. Which of the following is an expression of the acceleration of the system?
Three boxes tied by two ropes move across a frictionless surface pulled by a force
Tap to reveal answer
Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.
A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between and as . Similarly, we will label the tension experienced by the rope between and as .
For the box on the right (of mass ) we know that the force pulls to the right and the rope pulls it to the left with a tension of .
We get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).
For the middle box (of mass ) we have a rope pulling on it on the right with tension and a rope pulling on the left with tension .
So we get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).
Finally, for the box on the left (of mass ) we have only one rope pulling it to the right with a tension .
So we get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).
Now it is just a matter of simple substitution. We have three equations:
From them we can get an expression for the force. Isolate in the second equation.
Substitute the expression of from the first equation and simplify.
Use this value of in the third equation to get an expression for the force.
So we have:
Solve for acceleration.
Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.
A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between
For the box on the right (of mass
We get the equation:
For the middle box (of mass
So we get the equation:
Finally, for the box on the left (of mass
So we get the equation:
Now it is just a matter of simple substitution. We have three equations:
From them we can get an expression for the force. Isolate
Substitute the expression of
Use this value of
So we have:
Solve for acceleration.
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A box is being pushed against a wall by a force as shown in the picture. If the coefficient of static friction between the box and the wall is , which of the following expressions represents what the force must be for the box not to fall?
A box is being pushed against a wall by a force
Tap to reveal answer
This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force . The box experiences the downward force of its own weight given by . Finally, since the box is trying to slide down, friction opposes this motion. The following diagram shows these forces:
The key is that the box should NOT move.
For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.
Now, on the vertical axis we have fiction and the object's weight:
Finally, we use the equation for friction force to solve the problem.
Substitute the weight, since it is equal to the force of friction.
Isolate the normal force.
Since the normal force is equal to the applied force, this is our final expression.
This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force
The key is that the box should NOT move.
For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.
Now, on the vertical axis we have fiction and the object's weight:
Finally, we use the equation for friction force to solve the problem.
Substitute the weight, since it is equal to the force of friction.
Isolate the normal force.
Since the normal force is equal to the applied force, this is our final expression.
← Didn't Know|Knew It →
A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?
A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?
Tap to reveal answer
You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.
We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.
We can solve for the horizontal component of the force using the applied force as the hypotenuse:
You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.
We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.
We can solve for the horizontal component of the force using the applied force as the hypotenuse:
← Didn't Know|Knew It →
A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?
A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?
Tap to reveal answer
When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.
In addition, the rocket experiences the downward force of its own weight. This is given by:
We know that .
We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.
We know that (since is directed upwards and is directed downwards).
Finally we solve for acceleration:
When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.
In addition, the rocket experiences the downward force of its own weight. This is given by:
We know that
We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.
We know that
Finally we solve for acceleration:
← Didn't Know|Knew It →
Three boxes tied by two ropes move across a frictionless surface pulled by a force as shown in the diagram. Which of the following is an expression of the acceleration of the system?
Three boxes tied by two ropes move across a frictionless surface pulled by a force
Tap to reveal answer
Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.
A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between and as . Similarly, we will label the tension experienced by the rope between and as .
For the box on the right (of mass ) we know that the force pulls to the right and the rope pulls it to the left with a tension of .
We get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).
For the middle box (of mass ) we have a rope pulling on it on the right with tension and a rope pulling on the left with tension .
So we get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).
Finally, for the box on the left (of mass ) we have only one rope pulling it to the right with a tension .
So we get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).
Now it is just a matter of simple substitution. We have three equations:
From them we can get an expression for the force. Isolate in the second equation.
Substitute the expression of from the first equation and simplify.
Use this value of in the third equation to get an expression for the force.
So we have:
Solve for acceleration.
Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.
A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between
For the box on the right (of mass
We get the equation:
For the middle box (of mass
So we get the equation:
Finally, for the box on the left (of mass
So we get the equation:
Now it is just a matter of simple substitution. We have three equations:
From them we can get an expression for the force. Isolate
Substitute the expression of
Use this value of
So we have:
Solve for acceleration.
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A box is being pushed against a wall by a force as shown in the picture. If the coefficient of static friction between the box and the wall is , which of the following expressions represents what the force must be for the box not to fall?
A box is being pushed against a wall by a force
Tap to reveal answer
This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force . The box experiences the downward force of its own weight given by . Finally, since the box is trying to slide down, friction opposes this motion. The following diagram shows these forces:
The key is that the box should NOT move.
For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.
Now, on the vertical axis we have fiction and the object's weight:
Finally, we use the equation for friction force to solve the problem.
Substitute the weight, since it is equal to the force of friction.
Isolate the normal force.
Since the normal force is equal to the applied force, this is our final expression.
This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force
The key is that the box should NOT move.
For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.
Now, on the vertical axis we have fiction and the object's weight:
Finally, we use the equation for friction force to solve the problem.
Substitute the weight, since it is equal to the force of friction.
Isolate the normal force.
Since the normal force is equal to the applied force, this is our final expression.
← Didn't Know|Knew It →
A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?
A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?
Tap to reveal answer
You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.
We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.
We can solve for the horizontal component of the force using the applied force as the hypotenuse:
You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.
We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.
We can solve for the horizontal component of the force using the applied force as the hypotenuse:
← Didn't Know|Knew It →
A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?
A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?
Tap to reveal answer
When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.
In addition, the rocket experiences the downward force of its own weight. This is given by:
We know that .
We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.
We know that (since is directed upwards and is directed downwards).
Finally we solve for acceleration:
When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.
In addition, the rocket experiences the downward force of its own weight. This is given by:
We know that
We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.
We know that
Finally we solve for acceleration:
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Three boxes tied by two ropes move across a frictionless surface pulled by a force as shown in the diagram. Which of the following is an expression of the acceleration of the system?
Three boxes tied by two ropes move across a frictionless surface pulled by a force
Tap to reveal answer
Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.
A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between and as . Similarly, we will label the tension experienced by the rope between and as .
For the box on the right (of mass ) we know that the force pulls to the right and the rope pulls it to the left with a tension of .
We get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).
For the middle box (of mass ) we have a rope pulling on it on the right with tension and a rope pulling on the left with tension .
So we get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).
Finally, for the box on the left (of mass ) we have only one rope pulling it to the right with a tension .
So we get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).
Now it is just a matter of simple substitution. We have three equations:
From them we can get an expression for the force. Isolate in the second equation.
Substitute the expression of from the first equation and simplify.
Use this value of in the third equation to get an expression for the force.
So we have:
Solve for acceleration.
Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.
A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between
For the box on the right (of mass
We get the equation:
For the middle box (of mass
So we get the equation:
Finally, for the box on the left (of mass
So we get the equation:
Now it is just a matter of simple substitution. We have three equations:
From them we can get an expression for the force. Isolate
Substitute the expression of
Use this value of
So we have:
Solve for acceleration.
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A box is being pushed against a wall by a force as shown in the picture. If the coefficient of static friction between the box and the wall is , which of the following expressions represents what the force must be for the box not to fall?
A box is being pushed against a wall by a force
Tap to reveal answer
This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force . The box experiences the downward force of its own weight given by . Finally, since the box is trying to slide down, friction opposes this motion. The following diagram shows these forces:
The key is that the box should NOT move.
For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.
Now, on the vertical axis we have fiction and the object's weight:
Finally, we use the equation for friction force to solve the problem.
Substitute the weight, since it is equal to the force of friction.
Isolate the normal force.
Since the normal force is equal to the applied force, this is our final expression.
This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force
The key is that the box should NOT move.
For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.
Now, on the vertical axis we have fiction and the object's weight:
Finally, we use the equation for friction force to solve the problem.
Substitute the weight, since it is equal to the force of friction.
Isolate the normal force.
Since the normal force is equal to the applied force, this is our final expression.
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A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?
A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?
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You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.
We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.
We can solve for the horizontal component of the force using the applied force as the hypotenuse:
You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.
We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.
We can solve for the horizontal component of the force using the applied force as the hypotenuse:
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