A massless rope that passes over a frictionless pulley connects two blocks. The

Question

A massless rope that passes over a frictionless pulley connects two blocks. The mass of the block on the left is 10kg, and it rests on a plane making an angle of 40 degrees with respect to the horizontal. The block on the right rests on a plane making an angle of 30 degrees with respect to the horizontal.

a) What is the mass of the right side block if the system accelerates to the right at 1 m/s2?

b) If the mass of the right side block is 5 kg, find the acceleration of the system.
c) Calculate the tension force in the rope under the conditions of part a and b.

Explanation / Answer

a)

let mass in right be M

Here , Using second law of motion

a = Fnet/total mass

a = (M*sin(30) - 10*sin(40))*g/(M+10)

as a = 1

M + 10 = M*4.9 - 63

solving for M

M = 18.72 Kg

the mass of right block is 18.72 Kg

b)

For M = 5 Kg

a = ( - 5*sin(30) + 10*sin(40))*g/(5+10)

a = 2.57 m/s^2 towards left

c)

For case a)

for left block

T = 10*a + 10*g*sin(40)

for a =1

T = 73 N

tension is 73 N

for case b)

for right block

T = 5*a + 5*g*sin(30)

for a = 2.57 m/s^2

T = 37.4 N

tension in string is 37.4 N

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