Newton's second law states that:

\(\displaystyle F=ma\)

We are given the mass of the object and the acceleration. Using these values, we can solve for the necessary force.

\(\displaystyle F=5kg*3\frac{m}{s^{2}}\)
\(\displaystyle F=15N\)

If an object is moving with constant velocity, then its acceleration must be zero.

\(\displaystyle a=\frac{v_2-v_1}{t}\)

\(\displaystyle v_2=v_1\rightarrow a=\frac{v-v}{t}=\frac{0}{t}=0\)

We can then look at Newton's second law. If the acceleration is zero, then the net force must also be zero.

\(\displaystyle F=ma\)

\(\displaystyle F=m(0)=0N\)

This means that the gravitational force and normal force cancel out, and the propulsion force of the car cancels out the force of friction. Forces may still be acting in respective directions, but the net sum of these forces is zero.

First we need to find the net force, which will be equal to the sum of the forces on the bone.

\(\displaystyle F_{net}=F_1+F_2\)

Since the forces are going in opposite directions, we know that one force will be negative (since force is a vector). Conventionally, right is assigned a positive directional value. The force to the left will be negative.

\(\displaystyle F_{net}=22N+(-17N)\)

\(\displaystyle F_{net}=5N\)

From here we can use Newton's second law to expand the force and solve for the acceleration, using the mass of the bone.

\(\displaystyle F=ma\)

\(\displaystyle 5N=(0.07kg)a\)

\(\displaystyle \frac{5N}{0.07kg}=a\)

\(\displaystyle 71.4\frac{m}{s^2}=a\)

Newton's third law says that for every force, there is an equal and opposite force: for an object on a non-inclined plane, the force of gravity is straight down and the normal force is straight up.

(For non-inclined planes, \(\displaystyle F_N=-(\cos(\theta)*mg)\) still works, just use \(\displaystyle 0^\circ\) for your \(\displaystyle \theta\) as the cosine of \(\displaystyle 0^\circ\) is \(\displaystyle 1\).)

In non-mathematical terms, the normal force is what keeps us from sinking into the ground: we experience the force of gravity and yet we aren't sucked into the Earth's crust. Why? Because the object we stand upon provides an equal but opposite force on our feet.

This principle still applies to objects on an inclined plane. As the angle of the plane increases, the force of gravity is divided. Gravity causes the object to travel downward and horizontally along the plane's surface. The normal force must still counteract the downward portion of gravity, but will be less than the total gravitational force since the object is sliding horizontally. This requires us to use the cosine term to reduce the total normal force. Eventually, if the angle of incline increases enough, we would be taking \(\displaystyle cos(90^o)\), which is zero; if the object is in free fall (the plane is vertical), then there is no normal force.

The normal force on an object at rest on a flat surface is equal to the gravitational force on that object. In equation form, that means \(\displaystyle \vec{F}_{normal}=-\vec{F}_{gravity}\).

Use Newton's second law (\(\displaystyle \vec{F}=m\vec{a}\)) to solve for the gravitational force, using acceleration due to gravity as your \(\displaystyle \vec{a}\).

\(\displaystyle \vec{F}=m\vec{a}\)

\(\displaystyle \vec{F}_{gravity}=(22kg)(-9.8\frac{m}{s^2})\)

\(\displaystyle \vec{F}_{gravity}=-215.6N\)

As normal force is the opposite of the gravitational force, we can see that \(\displaystyle \vec{F}_{normal}=215.6N\).

Looking at our given values, we can see that each term only has one significant figure, since zeroes in front of the decimal are not significant. We can immediately determine that our answer will also have one significant figure however, to maintain accuracy, we will only round for precision after the final calculation.

The normal force acting on box is equal and opposite to the weight of the box. When the box is on a level surface, this term is equal to the force of gravity.

\(\displaystyle F_N=-F_g\)

\(\displaystyle F_g=mg\)

\(\displaystyle F_N=-mg\)

Use our given mass and the acceleration due to gravity to solve for the normal force of the box.

\(\displaystyle F_N = -(20kg)(-9.8\frac{m}{s^2}) = 196N\)

This value is equal to the total upward force on the box to counter its weight. When the man is lifting the box, he accommodates part of this upward force, lessening the normal force.

\(\displaystyle F_{up}=196N=F_{man}+F_{N_f}\)

Find the final normal force by subtracting the force from the man from the total upward force.

\(\displaystyle 196N - 50N = 146N\)

Our given numbers only allow for one significant figure, so we must round down.

\(\displaystyle 146N\approx 100N\)

Normal force is defined as the force that a surface exerts on an object. If the object is at rest, net force on the object is equal to zero; therefore, the downward force (weight) must be equal to the upward force (normal force).

\(\displaystyle F_{net}=F_g+F_N\)

Since weight acts in the downward direction it will be negative. The total sum of the forces must be zero, in order for the object to be at rest.

\(\displaystyle 0N=(-10N)+F_N\)

\(\displaystyle 10N=F_N\)

This is also in accordance with Newton's third law, which suggests that the normal force will be equal and opposite the force of weight.

Normal force is the force that the table pushes onto the block. If the force on the block from gravity is pushing the block downward, then the normal force has to push straight upward to keep the block at rest. Since the block is not moving, these two forces must be equal and opposite to produce a net force of zero.

\(\displaystyle F_N=F_g=mg\)

Since normal force is pushing upward, it will be positive. (The force of gravity would be negative).

 \(\displaystyle F_N=100kg*10\frac{m}{s^{2}} = 1000N\)

The correct answer is 43N. Since the box is on an incline, normal force balances with the component of gravity that is perpendicular to the surface of the incline.

\(\displaystyle F_N=-F_{g_y}=-(mg\cos\theta)\)

Note that the normal force is in the upward (positive) direction, while gravitational acceleration and the force of gravity are in the downward (negative) direction.

We are given the weight of the box (note that it will be downward and, therefore, negative):

\(\displaystyle mg=-50N\)

Use this to solve for the normal force.

\(\displaystyle F_N=-(-50N\times\cos30^o)\)

\(\displaystyle F_N=43N\)

Normal force always acts perpendicular to the contact surface. It can act counter to gravity or to any other force that pushes an object against a surface.

Think of normal force as the force that keeps a surface solid; without normal force, an object would pass right through a surface. With this in mind, consider a book that you push against a wall in the horizontal direction such that it is in equilibrium and does not slide down the wall. In this case, normal force is acting counter to your pushing force along the horizontal axis and perpendicular to the force of gravity. Another example is any object placed on an inclined plane; in this system the normal force will act perpendicular to the incline and will only counter a portion of the gravitational force.