Question

a light board that is 10 m long is supported by two sawhorses, one at one end of the board and a second at the midpoint. a 40-n weight is placed between the two sawhorses, 3.0 m from the end and 2.0 m from the center. what is the magnitude of the force the end sawhorse exert on the board?

Answer 1

Using the principles of static equilibrium and selecting the midpoint sawhorse as the pivot, the force exerted by the end sawhorse on the board is calculated to be 16 newtons upward.

Step-by-step explanation:

We can solve this physics problem using the principles of static equilibrium.

The board is in static equilibrium, which means the sum of all vertical forces and the sum of all torques about any point must be zero.

To find the force exerted by the end sawhorse, we'll set up the problem like this:

• The weight of the board is negligible.
• The weight acts downward 3.0 m from the end, hence 1.0 m from the midpoint.
• The two sawhorses act as supports at 0 m and 5 m from one end.

Now, we choose the midpoint sawhorse as the pivot point, as its force won't contribute to torque about its own location. We set the sum of torques around the midpoint to zero:

torque = 0 = (Force at end * 5 m) - (40 N * 2 m)

Solving for the force at the end, we get:

Force at end = (40 N * 2 m) / 5 m

= 16 N

Therefore, the force that the end sawhorse exerts on the board is 16 newtons upward.