Two blocks and pulley?

Question

A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction µ = 0.75. A string attached to the block passes over a pulley to a block of mass m3 = 4 kg. The pulley is a uniform disk of mass m2 = 0.4 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

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a) With what acceleration does the mass m3 fall?

a = m/s2

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b) What is the tension in the horizontal string, T1?

T1 = N

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c) What is the tension in the vertical string, T3?

T3 = N

Answer 1

Let:

a be the acceleration of the masses m1 and m3,

u be the coefficient of friction,

R be the normal reaction of the table on m1,

g be the acceleration due to gravity.

For the mass m1:

T1 - uR = m1 a

m1 g = R

Eliminating R:

T1 = m1 (a + ug) ...(1)

For the mass m3:

m3 g - T3 = m3 a

T3 = m3(g - a) ...(2)

For the pulley with radius r, moment of inertia m2 r^2 / 2, and angular acceleration alpha:

r (T3 - T1) = m2 r^2 alpha / 2

a = r alpha

Eliminating alpha:

T3 - T1 = m2 a / 2 ...(3)

(a)

Substituting in (3) for T1 from (1) and T3 from (2):

m3 (g - a) - m1 (a + ug) = m2 a / 2

2m3 (g - a) - 2m1 (a + ug) = m2 a

(2m1 + m2 + 2m3)a = 2(m3 - u m1)g

a = 2(m3 - u m1)g / (2m1 + m2 + 2m3)

= 2( 4 - 0.75 * 2)g / (2 * 2 + 0.4 + 2 * 4)

= 5 * 9.81 / 12.4

= 3.9556 -> 3.96 m/s^2.

(b)

From (1):

T1 = 2(3.9556 + 0.75 * 9.81)

= 22.626 -> 22.6 N.

(c)

From (2):

T3 = 4(9.81 - 3.9556)

= 23.417 -> 23.4 N.