Some examples of physics problems demonstrating the concept of Torque

Example Ladder Problem
A 10 meter long ladder leans against the wall as shown. If the ladder weighs 100N, what is minimum mu?

In order to begin solving this problem, you must use the given length of the ladder and measure of the angle to solve both the larger and smaller triangles. By either using simple trigonometric functions such as sine or cos, or recognizing that both triangles are 3-4-5 triangles, you must find the measure of the vertical and horizontal components. Before you can find the friction force and then the minimum mu, you need to find the force of the wall. To do this, you need to set the torques (both sets of coupling forces) equal to each other.

(Fwall)(Moment Arm Wall) = (Fladder)(Moment Arm Ladder)

(x)(8) = (100)(3)

You must multiply x (the force of the wall) by its vertical component, which is eight. This is then set equal to 100 times 3, which represent the weight of the ladder (100) times the horizontal component of the ladder, which is three.

x = 300/8

x = 37.5 The force of the wall is equal to 37.5 newtons.

Since the force of the wall and friction are couple forces, they are congruent so the friction force is also 37.5 newtons.

To find the mu, use the equation (Ffriction) / (Fnormal) = mu

so (37.5) / (100) = 0.375 = mu

The minimum mu needed is 0.375.

See-Saw Problem

Given: Weight of the Board (WB) = 800 N, Weight of Person 1 (W1) = 1500 N, Weight of Person 2 (W2) = 2000 N

Find: The Force Provided by Each Support

To begin this problem, one must pick a support to use as the fulcrum. Let's say that the leftmost support is the fulcrum (called S1 and other support is S2). Because the system is in equilibrium, the torques are balanced, so if we set the torques acting clockwise (W1 and S2) equal to the torques acting anti-clockwise (WB and W2) we can calculate how force S2 supports. To find the torque, one must multiply the force exerted from the object (mass of the objects in this case) by its distance from the lever arm.

W1(2) + S2(2) = WB(1.5) + W2(5)

1500(2) + S2(2) = 800(1.5) + 2000(5)

S2 = 4100 N

This answer gives you the force exerted by support 2. Now that you know the force of one of the supports, there are two approaches to solving for support 1. You could either do the same thing using support 2 as your fulcrum and calculate support 1 that way or you could set the forces acting downward (W1, WB, W2) equal to the forces acting upward (S1 and S2). This can be done because the system is in equilibrium, the board is not moving up or down and is not rotating, meaning that the forces acting up are in balance with forces acting down.

W1 + WB + W2 = S1 + S2

1500 + 800 + 2000 = S1 + 4100

S1 = 200