A 6.00 meter long board was placed on two sawhorses. The left end of the board w

Question

A 6.00 meter long board was placed on two sawhorses. The left end of the board was
1.50 meters from one of the sawhorses. The right end of the board was 2.50 meters from the other sawhorse. The board’s mass was 50.0 kg and its center of mass was 2.50 meters from the left end of the board. A 750. newton mass was placed 0.250 meters to the right of the sawhorse on the right.
a) Calculate the upward force exerted by the sawhorse on the left.
b) Calculate the upward force exerted by the sawhorse on the right.

Explanation / Answer

---------------1.5-----------2.5--------------------4.5---------------6
____________________________________________________
   H Center H

Draw a picture to help determine the positions.

The first sawhorse is at position (x) = 1.5 m

The second sawhorse is at x=4.5 m

The 750N mass is placed at x= 4.75 m.

The center of mass of the board is at x=2.5 m

This is a torque problem. So, multiply the force from the mass by the perpendicular distance to the sawhorse. There is also a force at the center of the board caused by its own weight : F=50 kg*9.8 m/s^2. Do the same thing with this.

The force on the sawhorse on the right is:

F=(750 N)(4.75-4.5 m) + (50 kg)(9.8 m/s^2)(2.5-4.5 m)

= (750 N) (0.25 m) + (50 kg)(9.8 m/s^2)(-2.0 m)

= -792.5 Nm

The force on the sawhorse on the left is:

F= (750 N) (4.75-1.5 m) + (50 kg)(9.8 m/s^2)(2.5-1.5 m)

= (750 N) (3.25 m) + (50 kg)(9.8 m/s^2)(1.0 m)

= 2927.5 Nm

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