in the arrangement as shown tension t2 is (g=10m/S2)
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Answered by
130
Given :
m=10kg
g=10m/s2
When a body of mass m is suspended by two strings making an angle α and β with horizontal then:
T1=mgcosβ/sin(α+β)
T2=mgcosα/sin(α+β)
β=30
α=60
T2=10x10xcos(60)/sin(60+30)
=100x(1/2) / 1=50N
∴Tension T2=50N
m=10kg
g=10m/s2
When a body of mass m is suspended by two strings making an angle α and β with horizontal then:
T1=mgcosβ/sin(α+β)
T2=mgcosα/sin(α+β)
β=30
α=60
T2=10x10xcos(60)/sin(60+30)
=100x(1/2) / 1=50N
∴Tension T2=50N
Answered by
98
The pulley is at rest . Therefore, the overall force from the block of 10kg will be 2 x 10 x g = 200N.(assume it to be T3) and not 100N.
According to lami's theorem, we know that :
T1/sin120 = T2/sin150 = T3/sin90.
Hence , 200/sin90 = T2/sin150
T2 = 200 x 1/2 x 1 = 100N
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