Question

1. A 3.0 meters long solid shaft makes an angle of 35 degrees with the horizontal. A vertical

force of 50 N is applied 0.60 m from the upper end. What is the magnitude of the torque

due to the vertical force about each end?

2. A Ladder of 6.0 meters in length weighs 150 N rests on horizontal ground and leans at an

angle of 65 degrees with the horizontal against a smooth vertical wall. How far up the ladder

may a 700 N person go before the ladder slips? The coefficient of friction between ladder

and ground is 0.40.

3. A Flat bar is loaded with boxes as shown in

figure at right. Box M = 50 kg, what must be

the magnitude of mass m so that the bar

becomes level? Neglect the weight of the bar.

What torque about each end of the bar?

Answer 1

1) The torques from the vertical force around the upper end is the lever times vertical component of the force:

$$τ _ u ​ =−F \cosθ⋅l=50\cos35°⋅0.6=−24.6\ Nm.$$

Around the lower end:

$$τ _ l ​ =F \cosθ⋅(L−l)=50 \cos35°(3−0.6)\\= 98.3\ Nm.$$

3) If the length of the bar is L, and its axis of rotation is at distance d from the point where mass M is attached, the value of mass m in order to balance M can be found from the equilibrium of torques:

$$Mgd=mg(L−d),$$

The torque about each end of the bar is

$$τ _ M ​ =Md,\\ τ _ m ​ =m(L−d).$$