Physics 133
Spring Quarter 2004
5 April 2004.
Email the quiz back to me at aubrecht@mps.ohio-state.edu.
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The quiz is due before class today, 5 April 2004.
The 40.0 kg uniform bar is 1.85 m long. This bar is to be balanced at a point that is 62.0 cm from the left end of the bar by a weight W that is to be placed at a point that is 14.0 cm from the left end of the bar.
a. What is the weight W? (5)
b. What is the force that is exerted at the balance point? (5)
take me to the journal assignments
Gordon’s solution will appear here
a. The sum of the torques would be given as
+(0.62 m - 0.14 m)(W) - (0.925 m - 0.62 m)(40 kg)(10 m/s2) = 0.
This comes about because the point of application of the weight W is 0.48 m to the left of the fulcrum, while the CM at 0.925 m is 0.305 m to the right of the fulcrum.
Solving,
W = (0.305 m)(400 N)/0.48 m = 254 N.
b. The sum of the forces must be zero also if it is to be in equilibrium, so the net downward force on the balance point is 400 N + 254 N = 654 N, and the net upward force on the balance point is 654 N, adding to zero.