Physics Levers Question

Question

A uniform bar AB is 2m long and weighs 120N. An upward force of 50N is applied 40cm from end A. A downward force of 60N is applied at end A and a downward force of 80N is applied at end B. Find the magnitude, direction, and location of the equilibrant force.

Answer 1

In order for the beam to be in equilibrium, all of the forces and all of the torques must be balanced. In order to balance the forces, we just need to add up all the upward forces and all the downward forces. The only upward force given is 50N. The three downward forces (weight is a downward force) sum to 260N. Thus, there must be an upward force of 210N in order for the beam to be in equilibrium.

Of course this is not enough, the torques must also be balanced. This is a little trickier because torque depends on the location of the applied force. We must first pick an origin, a convenient origin is the end of the beam labeled A. We will measure all distances from this point. The torque associated with an applied force is given by T = F*d*cosθ. In our case, all angles are right angles, so cosθ will always equal 1 and we can go with T = F*d. Here, d represents the distance from the origin to the location where the force is applied.

Now, in order to balance the torques, we must sum the clockwise torques and the counterclockwise torques and make sure they are equal. The clockwise torques will be those associated with the downward forces:

(60N)(0m) + (120N)(1m) + (80N)(2m) = 280Nm. Recall that the weight can be treated as acting on the center of the beam. The counterclockwise torques are: (50N)(0.4m) + (210N)(x). Here x is the unknown location of the 210N force. This gives us the equation:

20Nm +(210N)x = 280Nm

which means x is about 1.24m.

Thus, we need a 210N upward force 1.24m from A in order to get equilibrium.