Problem 10.42 The boom in the figure below (Figure 1)weighs 2100 N and is attach

Question

Problem 10.42 The boom in the figure below (Figure 1)weighs 2100 N and is attached to a frictionless pivot at its lower end. It is not uniform the distance of its center of gravity from the pivot is 34 of its length. Figure 1 Guy wire Boom 60.0° 5000 N Make a free-body diagram of the boom. The boom attaches to the building at dot A. The center of gravity of the boom is indicated by dot B. The guy wire attaches to the boom and holds the load att dot C. All force vectors should begin at the center of a black dots. The orientation of your vectors will be graded. Positive torque is considered to be in the counterclockwise direction. add element E vector sum Xa element i attributes Ureset 2 hep delote Guy wire Boom ulload Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining

Explanation / Answer

(A) use torques:

torque from guy wire = torque from gravity + torque from hanging weight

T * L sin60 = 2100 * 0.34L * sin30 + 5000 * L * sin30

T sin60 = 357 + 2500

T = 3298.98 Newtons

(B) only horizontal forces are Hh and guy wire. They must balance. So

Hh = 3298.98 Newtons

(C) only vertical forces are from Hv and weights so

Hv = 2100 + 5000 = 7100 Newtons

According to Chegg guidelines I can answer one question at a time so please post other questions separately ...

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