Torque EX 3

The diagram shows the gravitational force acting on the ladder, the pivot (black dot) and the lever arm. The magnitude of the gravitational force is 70 N. The gravitational force acts at the center of mass of the ladder which is located 5 feet = 5 × 0.3 = 1.5 m from the pivot point (one end of the ladder). Therefore, the magnitude of the lever arm is 1.5 m.

To solve this problem we first move the two vectors ${\vec {r}}$ and ${\vec {F}}_{G}$ so that they are arranged "tail to tail"; we then draw the x-axis parallel to the lever arm and we determine the angle between them:

Thus we have to determine the angle $\alpha$. But from the triangle we see that the side adjacent to $\alpha$ is 3 feet = 3 × 0.3 = 0.9 m and the hypotenuse is the length of the ladder or 10 feet = 10 × 0.3 = 3 m. The angle $\alpha$ is:

$\alpha =\arccos({0.9 \over 3})=72.5^{\circ }$

From the diagram we see that the angle that we will use in the equation for the torque is 72.5° + 90° = 162.5°.

Since we drew the vector of the force in the "third quadrant" we conclude that the direction of the torque is clockwise and therefore the torque is negative:

$\tau =-F_{G}r\sin {\theta }=-(70)(1.5)\sin {162.5^{\circ }}=-31.6Nm$