Equilibrium of a Rigid Body A horizontal uniform bar of mass 2.5 kg and length 3
ID: 1473445 • Letter: E
Question
Equilibrium of a Rigid Body
A horizontal uniform bar of mass 2.5 kg 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 0.6 m from the other end. A monkey of mass 1.25 kg 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.
N
Part D
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 newtons using three significant figures.
T1 = 13.8N
Explanation / Answer
M =2.5 kg , L = 3 m
m =1.25 kg
Net torque about on axis passing through strin 2 is zero.
Torque =rXF
(1.25*9.8)*(3/2 - 0.6) +(2.5*9.8)*(3/2 -0.6) = 2.4T1
T1 = 13.8 N
Net torque about on axis passing through strin 1 is zero.
Torque =rXF
(1.25*9.8)*(3/2) +(2.5*9.8)*(3/2 ) = (3-0.6)T2
T2 = 22.97 N
Another way
Net force is along vertical direction is zero
T1+T2 =Mg +mg
T2 = (2.5+1.25)*9.8 -13.8
T2 = 22.97 N
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.