Question

A 28.0-kg block is connected to an empty 2.56-kgbucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.41 and the coefficient of kinetic friction between the table and the block is 0.32. Sand is gradually added to the bucket until the system just begins to move. Ignore mass of cord. (Figure 1)

Calculate the mass of sand added to the bucket.

Calculate the acceleration of the system.

Answer 1

Mass of the block (M) =28 kg
mass of the bucket (m) = 2.56 kg
(a)
Now consider the mass of the sand is m_s , when it begins to move
f_s is friction force
Now consider FBD of the bucket
T = (m+m_s)g ---------(1)
Now consider the FBD of mass M
T = f_s
f_s = u_S*R = u_s*Mg = 0.41*(28*9.81) = 112.62 N
T = 112.62
now putting this in equation 1
112.62 = (2.56+m_s)*9.81
On solving we get
m_s = 8.92 kg
Hence it will slide when mass of the sand become 8.92 kg .
(b) Now when it start to move then the friction will be kinetic.
now considering the FBD of bucket
(m+m_s)*g - T = (m+m_s)a
(2.56+8.92)*9.81 - T = (2.56+8.92)a
112.62 - T = 11.48a
T = 112.62 - 11.48a -------------(1)
Now considering the FBD of mass M
T - f_k = Ma
T - (u_k*Mg) = Ma
T = Ma + (u_k*Mg)
T = 28a +(0.32*28*9.81)
T = 28a + 87.898 ---------(2)
equating 1 and 2
28a + 87.898 = 112.62 - 11.48a
(28+11.48)a = 112.62 - 87.898
a = 0.626 m/s²
hence the acceleration of the system will be 0.626 m/s²