A 50N diver stands at the end of a 4.0m diving board. The board is attached by two supports 1.5 m apart as shown below. Find the tension in each of the two supports if the diving board weighs 150N.
Answers
Answer:
We take the force of the left pedestal to be F1 at x = 0, where the x axis is along the diving board. We take the force of the right pedestal to be F2 and denote its position as x=d. W is the weight of the diver, located at x=L. The following two equations result from setting the sum of forces equal to zero (with upward positive), and the sum of torques (about x2) equal to zero:
F1+F2−W=0
F1d+W(L−d)=0
(a) The second equation gives
F1=−dL−dW=−(1.5m3.0m)(580N)=−1160N
which should be rounded off to F1=−1.2×103 N . Thus, ∣F1∣=1.2×103 N
(b) F1 is negative, indicating that this force is downward.
(c) The first equation gives F2=W−F1=580N+1160N=1740 N
which should be rounded off to F1=1.7×103 N . Thus, ∣F1∣=1.7×103 N
(d) The result is positive, indicating that this force is upward.
(e) The force of the diving board on the left pedestal is upward (opposite to the force of the pedestal on the diving board), so this pedestal is being stretched.
(f) The force of the diving board on the right pedestal is downward, so this pedestal is being compressed.