WORLD FOR CIVIL

Example 1 - Torque review problems

In this example, two painters are standing on a 300 lb. scaffolding (beam) which is 12 ft. long. One painter weighs 160 lb. and the second painter weighs 140 lb. The scaffolding is supported by two cables, one at each end. As they paint, the painters begin wondering what force (tension) is in each cable.

The question is, what is the force (tension) in each cable when the painters are standing in the positions shown. Notice that for a uniform beam or bar, as far as equilibrium conditions are concerned, the beam weight may be considered to act at the center (of mass) of the bar.

We proceed with this problem with a well establish statics technique.

1. Draw a Free Body Diagram (FBD) showing and labeling all external forces acting on the structure, and including a coordinate system.
Notice in the diagram to the right, we have shown the forces in the cables supporting the beam as arrows upward, and labeled these forces T_A and T_B.

2. Resolve all forces into x and y-components. In this example, all the forces are already acting in the y-direction only, so nothing more needs to be done.
3. Apply the (2-dimensional) equilibrium conditions:

(No external x-forces acting on the structure, so this equation gives no information.) T_A - 300 lb - 160 lb - 140 lb + T_E = 0
Here we have summed the external y-forces. We can't solve this equation yet, as there are two unknowns (T_A and T_E) and only one equation so far. We obtain our second equation from the 2^nd condition of equilibrium - sum of torque must equal zero.

Before we can sum torque we must first PICK A POINT, as we always calculate torque with respect to a point (or axis). Any point on (or off) the structure will work, however some points result in an easier equation(s) to solve. As an example, if we sum torque with respect to point E, we notice that unknown force T_E acts through point E, and, if a force acts through a point, that force does not produce a torque with respect to the point (since the perpendicular distance is zero). Thus summing torque about point E will result in an equation with only one unknown, as shown
or -T_A(12) +300 lb(6ft) + 160 lb(3 ft) + 140 lb(1 ft) = 0

Where we determined each term by looking at the forces acting on the beam one by one, and calculating the torque produced by each force with respect to the chosen point E from: Torque = Force x perpendicular distance (from the point E to the line of action of the force). For a review of torque, select Torque.

The sign of the torque is determined by considering which way the torque would cause the beam to rotate, if the beam were actually pinned at the chosen point. That is, if we look at the 160 lb. weight of painter one (and ignore the other forces), the 160 lb. weight would cause the beam to start rotating counterclockwise (+), if the beam were pinned at point E.

It is important to note that the sign of the torque depends on the direction of the rotation it would produce with respect to the chosen pivot point, not on the direction of the force. That is, the 160 lb. force is a negative force (downward) in the sum of forces equation, but it produces a positive torque with respect to point E in the sum of torque equation. See equations below:
or, T_A - 300 lb - 160 lb -140 lb + T_E = 0
or, -T_A(12) +300 lb(6ft) + 160 lb(3 ft) + 140 lb(1 ft) = 0

We now solve the torque equation for T_A, finding T_A = 201.67 lb (round off to 202 lb). We then place the value for T_A back into the y-force equation and find the value of T_E = 398 lb.

We have now found the forces in the cables when the painters are in the positions shown in the problem. As an additional thought problem, one might consider the question of what is the minimum cable strength required so that the painters could move anywhere on the scaffolding safely. (Answer = 450 lb.)