Question 11, chap 112, sect 2 part 1 of 3 10 points A 1363.0 N uniform boom of l

Question

Question 11, chap 112, sect 2 part 1 of 3 10 points A 1363.0 N uniform boom of length is supported by a cable, as shown. The boom is pivoted at the bottom, the cable is attached a distance l from the pivot, and a 3283.0 N weight hangs from the boom's top. 32° 3283 N 58° Find the force Fr applied by the supporting cable Answer in units of N Question 12, chap 112, sect 2. part 2 of 3 10 points Find the horizontal component of the reac- tion force on the bottom of the boom Answer in units of N. Questio 13, chap 112, sect 2.

Explanation / Answer

Let the tension in cable be T

11) Let us consider moment equilibrium of forces about the pivot

3283*lcos58 = Tcos32*0.75lsin58 + Tsin32*0.75lcos58

T = 2319.6 N

Trension in cable is 2319.6 N

12) Horizontal component of reaction at bottom = Tcos32=2319.6*cos32=1967.2 N (towards right)

13)Verticla reaction at bottom = 3283-Tsin32 = 2053.8 N(upwards)

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