Force Equilibrium - MCAT Physical

The skydiver’s velocity (along the y-axis) will be zero when he initially jumps out of the plane, however, his net force will be greatest at that point.

As the skydiver first jumps, there is only the force of gravity acting on him. As he approaches terminal velocity, however, the force of wind resistance becomes equal and opposite to the force of gravity, and he begins to move at constant velocity with a net force of zero.

Consider that tension is a contact force that acts over a distance. Assuming the rope does not stretch and has no mass (as we should for the MCAT), we can think of tension as a force that acts directly on the mass. We can draw a force diagram below.

If the system is stationary (at equilibrium), we can see that tension is equal to the weight, T = mg.

T = (2 kg)(9.8 m/s2) = 19.6N

Given that the system is motionless and that the tensional force is constant throughout the rope, we know that the weights on the platform must balance the weight of the mass. Given that the mass weights 2kg and each weight is 0.5kg, we can determine the number of weights required by dividing 2kg/0.5kg = 4 weights. Note that we can neglect the weight of the pulley in this problem; while this is usually the case on the MCAT as well, be sure to double check.

Use Newton's second law, , to relate the forces and accelerations of the particles. Writing this out for the two charges gives the following equations.

The two charges exert equal-magnitude and opposite-direction forces on one another according to Newton's third law, so , or using the right side of the second law equations, .

Plugging in our given values allows us to solve for .

The velocity of the mass will be the greatest at the equilibrium position. Although the restoring force on the mass due to the spring will be the greatest at maximum displacement from the equilibrium position, this force is being applied to the mass as it moves in the opposite direction. As the mass moves towards the equilibrium position from maximum displacement, the restoring force due to the spring is applied in the same direction as the velocity of the mass. Because of this, the object is accelerating. After the mass passes through the equilibrium position, the spring begins to apply a force in the direction opposite to the motion of the mass, resulting in the deceleration of the mass. The mass experiences its maximum velocity at the equilibrium position.

When an object is at terminal velocity, the upward force on the object from air resistance is equal to the force of gravity. These equal, but opposing, forces cancel each other out and the net force is zero. This results in a zero acceleration, allowing the object to fall at a constant (terminal) velocity.

The man has reached a terminal velocity, which means that he has an acceleration of Because , the total force on the skydiver is 0N. At this point, the force due to gravity in the downward direction is equal to the force of air resistance in the upward direction.

In order for the potted plant to stay still, the net force on the plant must be 0N. The tension of the rope is the force that allows an object to hang without falling. In order to keep the plant stationary, the tension must be equal to 25N in the upward direction.

There are only two forces acting on the box: the force of gravity and the force of tension in the rope. Since the box is not in motion, we can assume that the system is in equilibrium and the net force is equal to zero.

Rearranging, we can see that gravity and the force of tension will be equal and opposite.

Calculate the force of gravity using Newton's second law.

Note that the force of gravity is negative, since it acts in the downward direction. Use this value to solve for the force of tension.

The best way to think of a system in equilibrium is that it is displaying a constant velocity in both the horizontal and vertical directions. Both the skydiver and the tightrope walker have a constant velocity. The driver, however, is driving around a circular track. You must remember that velocity is a vector of magnitude and direction. Because the direction of the driver is changing, he is not driving at a constant velocity; thus, he is not in equilibrium, and is experiencing an acceleration (centripetal acceleration).

Keep in mind that the skydiver is experiencing both the downward force of gravity, and the upward force of air resistence on his deployed parachute. These forces cancel, producing a net force of zero, and allowing him to fall at a constant velocity in equilibrium.

Since the only acceleration on the object is to the west, we can conclude that there is no vertical acceleration on the object. As a result, the northern force must be offset by the third force. The only way to offset the northern force is to have a force pointing south, in any direction. As such, the third force could be directly south, southwest, or southeast as long as the southern component can counteract the northern force.

The third force would only need to be equal to the northern force if it was pointing directly due south, and we can only confirm that the third force is not completely offsetting the western force. As a result, we can only conclude that the third force has some magnitude in the southern direction.

The motion of the boat () needs to counteract the motion of the man (). As the man's foot pushes him forward, the boat pushes back (Newton's third law). Due to opposing forces, the boat will move backward as the man moves forward.

This scenario can also be solved by looking at momentum. Both the man and the boat start from rest, so in order for momentum to be conserved once the man gains positive velocity, the boat must gain negative velocity.

When an object is decelerating, there is a net force in the opposite direction of the motion of the object. Usually, this force is friction, which always acts counter to the net velocity. Even if there is a force in the direction of the object's motion, the net force is backwards if the object is decelerating.

Mathematically, we can prove this concept using Newton's second law. The direction of the object's motion will be positive, while the opposite (backwards) direction will be negative.

If the object is decelerating, then the acceleration is negative.

From this calculation, the force must be negative, acting counter to the direction of motion.

Newton's first law indicates that an object in motion will "want" to remain in motion at constant velocity unless it is acted on by a net force. In this case, Newton's first law is the only reasonable answer since in space, it is assumed there is no frictional or gravitational forces acting on the shuttle. Newton's second law can be described using the equation . Note that his first law is just a special case of the second law, where . Snell's law describes the interaction between waves and different media. Kirchhoff's laws involve charge conservation within circuits. Pascale's principle indicates that pressure exerted within a fluid is exerted equally in all directions.

Since the two vectors are perpendicular, they form a right triangle with the vector sum. Therefore, we simply use the Pythagorean theorem to solve for the sum.

The two vector components' directions determine the vector sum's direction as well. Since the components are north and east, the resultant vector is roughly northeast.

Since the block is moving to the right, we know that the frictional force is acting to the left. Also acting to the left is the gravitational force on the hanging block. Since it is a constant velocity of , we know that acceleration is zero and the downward force on the hanging block plus the frictional force is exactly equal to the force . Add the two forces.

For acceleration to be zero, the force needs to equal the sum of the frictional force on the block and the gravitational force on the hanging block.

To then accelerate the system, an additional force is needed.

This is the additional force required. Thus the total force required is:

The skydiver’s velocity (along the y-axis) will be zero when he initially jumps out of the plane, however, his net force will be greatest at that point.

As the skydiver first jumps, there is only the force of gravity acting on him. As he approaches terminal velocity, however, the force of wind resistance becomes equal and opposite to the force of gravity, and he begins to move at constant velocity with a net force of zero.

Consider that tension is a contact force that acts over a distance. Assuming the rope does not stretch and has no mass (as we should for the MCAT), we can think of tension as a force that acts directly on the mass. We can draw a force diagram below.

If the system is stationary (at equilibrium), we can see that tension is equal to the weight, T = mg.

T = (2 kg)(9.8 m/s2) = 19.6N

Given that the system is motionless and that the tensional force is constant throughout the rope, we know that the weights on the platform must balance the weight of the mass. Given that the mass weights 2kg and each weight is 0.5kg, we can determine the number of weights required by dividing 2kg/0.5kg = 4 weights. Note that we can neglect the weight of the pulley in this problem; while this is usually the case on the MCAT as well, be sure to double check.