For questions 12 and 13, consider a horizontal, uniform board of weight 125 N an

Question

For questions 12 and 13, consider a horizontal, uniform board of weight 125 N and length 4 m that is supported by vertical chains at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N

12. What is the tension in the left chain?
a. 250 N
b. 375 N
c. 500 N
d. 625 N
e. 875 N

13. How far from the left end of the board is the person sitting?
a. 0.4 m
b. 1.5 m
c. 2 m
d. 2.5 m
e. 3 m
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3. A radio station broadcasts on a carrier frequency that emits a signal with a wavelength of 3 m. The frequency of the signal is most nearly _______.
a. 1 MHz
b. 10 MHz
c. 100 MHz
d. 1000 MHz
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Explanation / Answer

This is a static torque problem. The platform must not be allowed to translate or rotate (accelerate in any way) since each end is fixed vertically.

Use downward = negative.
The weight of the board is -125N, the weight of the person is -500N, the tension in one chain is +250N. There is no acceleration, so the net force must be = 0 (always make zero your best friend).

250 -500-125 +Tension = 0
Tension = 375 N

The next part is a torque situation. The net torque must be zero, since the whole thing does not rotate.

Make believe that the board will pivot around the left chain anchor point if the torques are not equal. Use clockwise as positive. The pivot point is 0 meters (moment arm). Assume that the weight of the board is uniformly distributed and it acts as a force at the center of the length of the board.

Sum the moment arms:
375N (0m) +125N(2m) +500N (distance from left pivot) -250N (4m) = 0 (see? Zero is your friend!)

Distance = 1.5 m from the left anchor point.

C=n=3x10^8

n= 10^8 Hz = 100 MHz

 

12-B

13-B

14-C

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