A block of mass m 1 = 2 kg rests on a table with which it has a coefficient of f

Question

A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction

A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction mu = 0.77. 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 = 0.6 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley. With what acceleration does the mass m3 fall? What is the tension in the horizontal string, T1? What is the tension in the vertical string, T3?

Explanation / Answer

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.

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