Mastering Physics Solutions: Pulling Three Blocks

Pulling Three Blocks

Part A = 4.50N
Part B = 1.50N

Solutions Below:

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force F. The magnitude of the tension in the string between blocks B and C is T = 3.00N. Assume that each block has mass m = 0.400kg.

Part A

What is the magnitude F of the force?

Since blocks A and B are not moving with respect to each other, you can treat them as one larger “object.” This larger object has the same acceleration as either block A or block B alone. The advantage of such an approach is that the larger object has only one force acting on it (the tension of 3.00N in the rope). Therefore the magnitude of the tension (tension here is equivalent to force) is:

F = ma
T = ma
3.00 = 2m * a
3.00 = 2(0.400) * a
3.00 = 0.800a

Solving for a yields:

a = 3.75m/s2

Therefore:

F = ma
F = 3(0.400)a
F = 1.2a
F = 1.2(3.75)

F = 4.5N

If this is confusing (didn’t we just say that the force was 3.00 and solve using that??), remember that when we solved for the acceleration, we used the tension on the string between blocks B and C. If we want the force on all 3 blocks, we have to incorporate the mass of all 3 into our equation. That was only possible by first solving for the acceleration of blocks A and B. Since all 3 blocks have the same acceleration, we can thus extrapolate using a = 3.75m/s2 and arrive at a total force of 4.5N.

Part B

What is the tension TAB in the string between block A and block B?

We can solve this using the acceleration from Part A:

F = ma
T = ma
T = 0.400 * 3.75

T = 1.5N

In other words, if Block B is pulling Block A thus creating tension on the string between them, the force equals mass times acceleration for just the one block (Block B). In Part A we solved for the force pulling all three blocks, but Part B just asks for the force pulling Block A.

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