A block of mass MB = 12.0 kg sits on a frictionless horizontal plane. It is conn

Question

A block of mass MB = 12.0 kg sits on a frictionless horizontal plane. It is connected by a string at one end which goes at an angle 40.0 degrees to the horizontal passing over a massless pulley to a suspended weight of mass M1 = 5.40 kg. From the other end of the block a string runs horizontally over a frictionless massless pulley and suspends a weight of M2 = 2.10 kg as pictured. What is the initial instantaneous acceleration of the block, taking right for the positive directs? Acceleration: 1.66 m/s^2

Explanation / Answer

M2g - T2 = M2a ..................[1]

T1 - M1g = M1a ..................[2]

and T2 - T1cos(40) = (MB + M1 + M2)a ..............[3]

so, from [1]; T2 = 2.10[9.8 - a]

similarly, T1 = 5.4[9.8 + a]

use these in [3] to get:

2.1[9.8 - a] - 5.4[9.8 + a]cos40 = (12+2.1+5.4)a

=> 20.58 - 2.1a - 40.54 - 4.136a = 19.5a

=> a = - 0.775 m/s2i which means that the acceleration will be to the left.

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