Tweedledum and Tweedledee are carrying a uniform wooden board that is L = 3.00m

Question

Tweedledum and Tweedledee are carrying a uniform wooden board that is L = 3.00m long and has a mass M = 15.0kg . If Tweedledum applies an upward force of magnitude F1 = 60.0N at the left end of the board, at what point and with what magnitude F2 of force does Tweedledee have to lift for the board to be carried?

If Tweedledum applies a force of F1 at the left end of the board, at what distance d from the left end and with what magnitude of force F2 does Tweedledee have to lift for the board to be carried?

Explanation / Answer

F2= 87N d= 2.53M

Method:

F1+F2 = 15kg*9.8m/s/s = 147N

torques must balance; if we sum torques around the left end, the torque due to F1 is zero since it has no moment around the left end;

the torque due to the weight of the board is: 147N*1.5m = 220.5Nm

the torque due to F2 is F2*d where d is the distance from the left end where F2 is applied, so we have

F2* d = 220.5Nm

since F1=60, we know that:

F2=147N-60N=87N,

therefore,

d=220.5Nm/87N = 2.53m

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