A yo-yo of mass
M = 0.240 kg
and outer radius R that is 2.25 times greater than the radius of its axle r is in equilibrium if a mass m is suspended from its outer edge as shown in the figure below. Determine the tension in the two strings,
T1
and
T2
and the mass m.
T1 = | N |
T2 = | N |
m = | kg |
Sum forces in the vertical direction to zero
Let m = unknown mass
Let M = mass of YoYo
T₁ - (m + M)g = 0
T₁ = (m + M)g
Sum moments about the YoYo center to zero
T₁[r] - mg[2.25r] = 0
T₁[r] = mg[2.25r]
T₁ = 2.25mg
(m + M)g = 2.25mg
m + M = 2.25m
M = 1.25m
m = M / 1.25
m = 0.240 / 1.25
m = 0.192.
m = 0.192kg <====Answer
T₁ = (0.192+ 0.240)9.81
T₁ = 4.237 N <====Answer
T₂ = mg
T₂ = 0.192(9.81)
T₂ = 1.883 N <====Answer
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