Figure 5-5 shows a man sitting in a bosun\'s chair that dangles from a massless

Question

Figure 5-5 shows a man sitting in a bosun's chair that dangles from a massless rope, which runs over a massless, frictionless pulley and back down to the man's hand. The combined mass of man and chair is 79.3 kg. With what force magnitude must the man pull on the rope if he is to rise (a) with a constant velocity and (b) with an upward acceleration of 1.04 m/s2? (Hint: A free-body diagram can really help.) Problem continues below. If the rope on the right extends to the ground and is pulled by a co-worker, with what force magnitude must the co-worker pull for the man to rise (c) with a constant velocity and (d) with an upward acceleration of 1.04 m/s2? What is the magnitude of the force on the ceiling from the pulley system in (e) part a (f) part b, (g) part c, and (h) part d? If the rope on the right extends to the ground and is pulled by a co-worker, with what force magnitude must the co-worker pull for the man to rise (c) with a constant velocity and (d) with an upward acceleration of 1.04 m/s2? What is the magnitude of the force on the ceiling from the pulley system in (e) part a (f) part b, (g) part c, and (h) part d?

Explanation / Answer

for simplicityt let us assume that when themotion of the chair is upward then it is positive (a) when the man is pulling the fiorce with atension T the total upward force on the man and chair due to itstwo contact points with the rope is 2T so the newtions second law of motion leadsto 2T - m g = m a when a = 0 T = mg/2 = 79.3 * 9.8 / 2 = 388.57 N (b) when the acceleration is a = 1.04 m /s2 then 2T - m g = m a 2T = m[g+a] T = m[g+a ] /2 = 79.3*[9.8+1.04]/2 T = 429.806 N (c) if the man is not holding the rope there is only one conact point between the rope and the man and chair so from the newtons second law of motion weget T - m g = m a now a=0 so T = mg = 79.3* 9.8 T = 777.14 N (d) when the acceleration is a = 1.04 m /s2 then T = m[g+a] = 79.3 * [9.8+ 1.04]= 859.612 N (e) the rope comes into contact at the left edge abdthe ruight edge of the pulley producing a total downward force of F = 2 T =2* 388.57 = 777.14 N (f) in part (b) the downward force on theceiling has a magnitude 2 T = 2* 429.806 N = 859.612 N (g) in part (c) the downward force on the ceiling hasa magnitude 2 T = 2*777.14 N = 1554.28 N (h) in part (d) the downward force on theceiling has a magnitude 2 T = 2* 859.612 N = 1719.224 N

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.