The system shown in the figure below consists of a mass M = 3.7-kg block resting

Question

The system shown in the figure below consists of a mass M = 3.7-kg block resting on a frictionless horizontal ledge. This block is attached to a string that passes over a pulley, and the other end of the string is attached to a hanging m = 2.3-kg block. The pulley is a uniform disk of radius 8.0 cm and mass 0.60 kg. What is the acceleration of each block? What is the tension in the string? Draw extended free-body diagrams showing the forces acting on each of the masses and the pulley. Apply Newton's second law to the two blocks, and in rotational form to the pulley to obtain three equations in the unknowns. Set up a separate coordinate system for each object such that they all accelerate in the .positive, direction. The linear and angular accelerations of the objects are related to each other because the string does not go slack.

Explanation / Answer

To begin with, its a good habit to draw free-body diagrams if this confuses you.

(a) Consider the blocks as a single system (So we get to ignore tension):

so the vertical force acting on the system is the weight of the "2.3 kg" block
w= mg =2.3 x 9.8N = 22.54N

While only the "2.3-kg" block contributes to the weight, the weight acts on both blocks:

Thus: F=ma --> 22.54=(2.3+3.7)a --> a=3.7566 m/s2

As long as the string remains taut, this acceleration is for BOTH the blocks

(b) Now consider the blocks individually

For the "3.7kg" block: the resultant force = Tension (T) --> T=ma =3.7*3.7566 = 13.90 N

For the "2.3kg" block: the resultant force = w- tension (t) --> t= w -ma = m (g-a) =2.3*(9.8-3.7566) =13.90N

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