A uniform metre stick of mass 200 g is suspended from the ceiling thorough two vertical strings of equal lengths fixed at the ends. A small object of mass 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tensions in the two strings.
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As the system in the translational equilibrium T1+T2=2+0.2=2.2
T1+T2=2.2
The rod does not rotate so the net torque is zero
Finding torque about c
0.2*0.2+T1*0.5=T2*0.5
(T2-T2)*0.5=0.04
(T2-T1)=0.04/0.5
=4/100*10/5=0.08
2T2=2.28
T2=1.14N
T1=2.2-1.14
=1.16N
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Answer:
As the system in the translational equilibrium T1+T2=2+0.2=2.2
T1+T2=2.2
The rod does not rotate so the net torque is zero
Finding torque about c
0.2*0.2+T1*0.5=T2*0.5
(T2-T2)*0.5=0.04
(T2-T1)=0.04/0.5
=4/100*10/5=0.08
2T2=2.28
T2=1.14N
T1=2.2-1.14
=1.16N
Attachments:
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