Question

An Atwood’s machine consists of two masses, mA = 1.55 kg and mB = 3.59 kg which are connected by a massless inelastic cord that passes over a pulley. If the pulley has radius 0.03 m and moment of inertia 3.04 kg m2 about its axle, determine the magnitude of tension force on mB, F_{TB i}n N.

Answer 1

Solution) mA = 1.55 kg

mB = 3.59 kg

R = 0.03 m

I = 3.04 kg m^2

Tension force on mB , FTB = ?

Pulley can be considered as solid disk so it's moment of inertia

I = (1/2)(m)(R^2)

3.04 = (1/2)(m)(0.03^2)

m = (2×3.04)/(0.03^2)

m = 6755.55 kg

(mB)(g) - T1 = (mB)(a)

T1 = (mB)g - (mB)(a)

T2 - (mA)(g) = (mA)(a)

T2 = (mA)a + (mA)g

Net torque , R(T1) - R(T2) = (I)(angular acceleration)

Acceleration , a = R(angular acceleration)

Angular acceleration = (a)/(R)

R( T1 - T2 ) = (1/2)(m)(R^2)(a)/(R)

T1 - T2 = (m/2)

(mB)g - (mB)a - (mA)a - (mA)g = (m/2)(a)

(mB - mA)(g) = (mB + mA + 0.5m)a

a = (mB - mA)(g)/(mB + mA + 0.5m)

a = [(3.59 - 1.55)(9.8)]/(3.59+1.55+0.5×6755.55)

a = 0.0059 m/s^2

So on substituting

(mB)g - (mB)a = T1

T1 = 3.59×9.8 - 3.59×0.0059

T1 = 35.16 N

Therefore FTB = T1 = 35.16 N