Centripetal force can be found using the equation:

In this situation, the radius is the length of the string. We are given the linear velocity and mass, as well as the radius, allowing us to solve for the centripetal force.

For this problem, we use the centripetal force equation:

We are given the mass, radius or rotation, and the linear velocity. Using these values, we can find the centripetal force.

This problem requires us to add torques about the pivot point. In order for the seesaw to be balanced, the torque must be equal on each side of the pivot point.

Use the equation for torque in this equation.

The force of each object will be equal to the force of gravity.

Gravity can be canceled from each side of the equation. for simplicity.

Now we can use the mass of the rock and the mass of the child. The total length of the seesaw is two meters, and the child sits at one end. The child's distance from the center of the seesaw will be one meter.

Solve for the distance between the rock and the center of the seesaw.

The metal bar is in rotational equilibrium, meaning there is no net torque. Use the wall pivot as the rotation point to eliminate having to calculate torque due to the unknown wall force, since the wall exerts no torque around the pivot. This leaves the rope tension and the weight of the bar as the only torque generators.

 Calculate the torques of those forces around the wall pivot with a standard procedure:

1)  Draw in the force vector arrows on the beam at their points of application.

2)  Draw a line through the force arrows extending to the edges of the drawing. This is the “line of force.”

3)  Draw a line perpendicular to the line of force from the axis of rotation (the wall pivot, in this case) to the line of force at the point where it is nearest the axis of rotation. This is the “perpendicular lever arm.”

The torque is then the force applied times its perpendicular lever arm around the axis of rotation.

Consider, for a uniform bar, the weight being applied at the center. The closest approach of the weight’s line of force to the pivot point is the center of the bar; hence, half the bar’s length is the perpendicular lever arm. So the torque of the weight, , around the pivot point is:

The tension is applied at the end of the bar, at a  angle with respect to the bar. The perpendicular lever arm can be calculated trigonometrically:

We can then find the torque generated by the tension force of the cable:

These torques produce rotations in the opposite direction, which is why they can cancel out to cause static equilibrium. Using the convention that counterclockwise rotations (as viewed in the diagram, with the wall on the left) are positive, the case of static equilibrium gives:

For this configuration, the tension in the cable happens to equal the weight of the bar.

When given a question about the angle of a ramp, compare it to the extreme angles: 0^o and 90^o.

1. When the ramp has an angle of 0^o, the net force 0. The force due to gravity must equal the normal force; thus the normal force is at a maximum value.

2. When the angle of the ramp is 90^o, the full force of gravity is experienced by the box, and there is no normal force. The net force is equal to the force of gravity. Remember that , so then theta is 90^o, force of gravity is at a maximum.

When we decrease the angle of the ramp, we get closer to scenario 1. As a result, we can conclude that the normal force on the box increases, rather than decreases.

The normal force of the woman is measured by the scale. While accelerating upward, the scale should read a larger weight than when it is at rest. When the elevator is moving upward at a constant speed, the scale should read the same as when it is at rest. This is because the normal force is generated to counter the downward forces pushing against the floor. When the elevator is accelerating, there is a net upward force from the acceleration as well as the normal force to counter gravity. The normal force generates an upward acceleration. When moving at a constant speed, there is no upward acceleration and the normal force acts only to counter gravity. The normal force, and scale reading, will thus be greater during the period of acceleration.

To find the normal force on the incline, we use the relationship:

This provides the magnitude of the force of gravity in the direction perpendicular to the incline. Normal force will always act in the direction perpendicular to the surface, and in this case will be equal and opposite to the force of gravity. We then plug in the mass and gravitational acceleration to find the normal force on this block:

The normal force is generated as a result of a force against a solid surface. As per Newton's third law, the surface will exert an equal and opposite force on the object in contact. If an object is resting on a flat surface, then the normal force will be working to counter the weight of the object due to gravity.

The normal force acting on the bench with five players is equal and opposite to the total weight of the bench and players. Keep in mind that weight acts in the downward direction.

When the two players stand up, the new normal force is reduced.

The difference in the normal force is:

We could also have found this change by adding the weights of the two players who stood.

First, calculate the gravitational force acting on the rock.

The exerts a force of downward, meaning that if the person exerted at least , then he or she would have been able to lift it up. Instead, the person applied only . This means that the person needed to apply  of additional force to lift the rock.

We know that normal force on a flat surface is equal and opposite the force of gravity.

Then, find the friction force between the crate and the floor using the equation:

This force is the minimum force required to start moving the crate.