The two blocks in the figure(Figure 1) are connected by a massless rope that pas
ID: 1296658 • Letter: T
Question
The two blocks in the figure(Figure 1) are connected by a massless rope that passes over a pulley. The pulley is 14cm in diameter and has a mass of 2.4kg . As the pulley turns, friction at the axle exerts a torque of magnitude 0.53N?m .
The two blocks in the figure(Figure 1) are connected by a massless rope that passes over a pulley. The pulley is 14cm in diameter and has a mass of 2.4kg . As the pulley turns, friction at the axle exerts a torque of magnitude 0.53N?m . If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?Explanation / Answer
take:
m1 (be the left hand mass).= 4.0 kg
T1 (be the left hand tension)
m2 (be the right hand mass) = 2.0 kg
m3 (be the pulley mass) = 2.4 kg
T2 (be the right hand tension)
g (be the acceleration due to gravity) = 9.78 m/s^2
a (be the acceleration of the masses)
alpha (be the angular acceleration of the pulley)
r (be the radius of the pulley) = 0.07 m
F (be the friction) = 0.53 N.m
h (be the high for m1) = 1 m
t (be the time taken to reach the floor)
*mass m1:
m1 g - T1 = m1 a
T1 = m1(g - a)
*mass m2:
T2 - m2 g = m2 a
T2 = m2(g + a)
*the pulley:
(T1 - T2)r - F = I alpha
= (m3 r^2 / 2)(a / r)
T1 - T2 = F / r + m3 a / 2
*Eliminate T1 and T2:
m1(g - a) - m2(g + a) = F / r + m3 a / 2
(m1 - m2)g - (m1 + m2)a - m3 a / 2 - F / r
(m1 + m2 + m3 / 2)a = F / r + (m1 - m2)g
a = [ F / r + (m1 - m2)g ] / (m1 + m2 + m3 / 2)
h = at^2 / 2
t = sqrt(2h / a)
= sqrt{ h(2m1 + 2m2 + m3) / [ F / r + (m1 - m2)g ] }
so t = sqrt{ 1( 2 * 4 + 2 * 2 + 2.4) / [ 0.53 / 0.07 + (4 - 2)9.78 ] }
= 0.7277 sec
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