Question

A uniform horizontal beam of weight 481 N and length 3.32 m has two weights hanging from it. One weight of 381 N is located 0.8798 m from the left end; the other weight of 281 N is located 0.8798 m from the right end. What must be the magnitude of the one additional force on the beam that will produce equilibrium?

Answer 1

To find the magnitude of the additional force needed to produce equilibrium in a horizontal beam with two weights, calculate the counteracting torque required and solve for the force value.

Step-by-step explanation:

To achieve equilibrium for a horizontal beam with weights hanging from it, the sum of all the forces as well as the sum of the torque (moments) about any point must be zero. For the given problem, there are three forces applied on the beam: the weight of the beam (481 N) acting at its center (1.66 m from either end) and two weights hanging, one of 381 N located 0.8798 m from the left end, and the other of 281 N located 0.8798 m from the right end.

To calculate the magnitude of the additional force needed to keep the beam in equilibrium, we first need to ensure the torques are balanced. Torque is calculated as the force times the perpendicular distance to the pivot point. Choosing the left end of the beam as the pivot point:

• Torque due to the 381 N weight: 381 N × 0.8798 m (counterclockwise)
• Torque due to the 281 N weight: 281 N × (3.32 m - 0.8798 m) (clockwise)
• Torque due to the weight of the beam: 481 N × 1.66 m (clockwise)

Now, an additional upward force F at the right end will provide a counterclockwise torque. Therefore, F needs to generate sufficient torque to counteract the clockwise torques. The sum of the counterclockwise torques must equal the sum of the clockwise torques:

381 N × 0.8798 m = F × 3.32 m + 281 N × (3.32 m - 0.8798 m) + 481 N × 1.66 m.

Solving for F gives us the magnitude of the additional force needed for equilibrium.

Answer 2

1143 N at 1.59 m from the left end

Step-by-step explanation:

For the system to produce equilibrium, the total force and moment must be 0. Since the total weight downward is

481 + 381 + 281 = 1143 N

Therefore the magnitude of the force acting upward to balance this system must be the same of 1143 N

That alone is not enough, we also need the position of the force for the total moment to be 0.

Let x be the length from the this upward force to the left side. And let the left point be the point of reference for moment arm:

481 * 3.32/2 + 381 * 0.8798 + 281*(3.32 - 0.8798) - 1143*x = 0

x = (481*1.66 + 381 * 0.8798 + 281*2.4402)/1143 = 1.59m