How to Find Tension & Acceleration in a Pulley System
If you have two blocks in a pulley system and need to know their acceleration and the line's tension, the general strategy of computation is as follows. The acceleration of the two blocks will be determined first. Then note that, for either block, the tension, T, and its acceleration, a, are related; therefore, T can be solved for, having already solved for acceleration, a.
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Instructions
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One Hanging Block
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1
Solve for force.
Newton's second law relates force and acceleration: F=ma. Mass m is known; that's the total mass of the blocks, or M1+M2. Acceleration, a, is unknown at first. Force, F, is known because it's the gravitational force pulling M1 downward. It is M1---g, where g is gravitational acceleration.
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2
Solve for acceleration of the two blocks.
F=ma becomes M1---g = (M1+M2)---a, where g is the gravitational acceleration. Solving for acceleration a is trivial arithmetic.
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3
Incorporate friction between M2 and the table.
Then the gravitational force M1---g is working both to accelerate the two masses and to fight friction. Therefore, the friction term would appear on the right-hand side of the above equation, which would then be written as M1---g = (M1+M2)---a + M2---g---mu, where mu is M2's coefficient of friction.
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4
Calculate tension, T, by looking only at the forces on M1.
The only forces acting on M1 are the gravitational force (downward) and the tension, T (upward). The result of these forces is the acceleration, a, of M1. So by the second law, F=ma becomes M1---g-T = M1---a. Acceleration, a, has already been solved for, so solving for T is just arithmetic.
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5
Calculate tension, T, by looking only at the forces on M2.
The only lateral forces acting on M2 are T and friction. The result of these forces is the acceleration, a, of M2. So by the second law, F=ma becomes T - M2---g---mu = M2---a. Acceleration, a, has already been solved for (it is the same value for both M1 and M2, since they are connected), so solving for T is just arithmetic.
Two Hanging Blocks
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6
Solve for force.
The left-hand side of the equation F=ma above changes from M1---g to (M2-M1)---g, for M2>M1. The gravitational force on the heavier block determines the direction of acceleration, but the force on the smaller block counters it to some extent.
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7
Solve for acceleration of the two blocks.
The right-hand side of the formula F=ma in the one-block-hanging case hasn't changed. So Newton's second law, F=ma, is written as (M2-M1)---g = (M1+M2)---a. Acceleration, a, is solved for with simple arithmetic.
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8
Pick a block, then solve for tension, T, based on the forces on it.
This is the same calculation as was performed on the single hanging block in the earlier problem. For example, pick M2. The forces on M2 are gravity (downward) and tension, T (upward). Their result is the acceleration of M2. So F=ma becomes M2---g - T = M2---a. Acceleration, a, was found above, so solving for tension, T, is simple arithmetic.
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