PSS 11.1 Equilibrium of a Rigid Body Learning Goal: To practice Problem-Solving
ID: 1265591 • Letter: P
Question
PSS 11.1 Equilibrium of a Rigid Body Learning Goal: To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.3kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.6m from the other end. A monkey of mass 1.15kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar. Part C Find T1, the magnitude of the force of tension in string 1, at the moment that the monkey is halfway between the ends of the bar. Express your answer in newton's using three significant figures. For the bar to experience no net force, what must be the tension T2 in string 2 when the monkey is halfway between the ends of the bar? Express your answer in newton?s using three significant figures.Explanation / Answer
Equilbrium
Net Force = 0
T1 + T2 = (m1+m2)*g
T1 + T2 = (2.3+1.15)*9.81 = 33.84 N
Net Torque about centre of road = 0
T1*1.5 + 0 + 0 - T2*(1.5-0.6) = 0
1.5 T1 - 0.9 T2 = 0
T2 = 5 T1 / 3
T1 + 5 T1 / 3 = 33.84 N
T1 - 33.84*3/8 = 12.69 N = 12.7 N
T2 = 33.84 - 12.69 = 21.15 N
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Answers :
T1 = 12.7 N
T2 = 21.15 N
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