A 10.5 kg board 6.00 m long is supported by two sawhorses, one at each end. A 4.45 kg saw sits on the board 1.80 m from the left end. Find the upward force that the LEFT sawhorse exerts on the board.

Answer 1

For these type of static problems, you usually start with a torque balance. The torque balance applies everywhere that is in equilibrium, so you pick a spot where you can eliminate one of the forces (or if there is a complicated force - like a hinge) by making the lever arm zero. In this case we do a torque balance around the point where the right sawhorse is applying an upward force.

The ccl torques = the cl torques (ccl is counter clockwise)

F of left sawhorse x lever arm to right sawhorse = Weight of the board x lever arm from middle of the board (COM) to the right sawhorse and the weight of the saw x the lever arm to the right sawhorse or

F_L(6.0 m) = (10.5 kg)(9.8 N/kg)(3 m) + (4.45 kg)(9.8 N/kg)(6 - 1.8 m)

Now solve for F_L

If you wanted F_R, you would now do a vertical force balance (F_R+F_L = F_W,B+F_W,S)

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