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Two bodies of mass 590 grams and 𝑚 grams are attached to the ends of a light inextensible string passing over a smooth pulley fixed to the edge of a smooth horizontal table.
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The first body rests on the table, and the other hangs freely vertically below the pulley.
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If the tension in the string is 90,860 dynes, determine the acceleration of the system.
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Let’s first notice that we’ve been given mass of the bodies in grams and tension in the string in dynes.
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Now, a dyne is a unit of force that acts on a massive one gram and increases its velocity by one centimeter per second every second along the direction it acts.
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And so, for consistency, we’re going to convert 𝑔 into centimeters per square second.
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Well, there are 100 centimeters in a meter, so we multiply by 100.
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We find 𝑔 is 980 centimeters per square second.
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The next thing that we need to do when we’re dealing with pulley problems is to draw a sketch.
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We have two bodies; let’s call them A and B.
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A sits on a smooth horizontal table.
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And B hangs vertically below the pulley from the piece of string.
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We really need to consider all the forces that act on each of our bodies.
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Well, we have the force acting downwards.
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It’s mass times acceleration.
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And the acceleration here is due to gravity.
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So it’s 590 times 𝑔.
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There is a reaction force acting in the opposite direction.
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And then on body A, we also have the tension in the string.
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The same amount of tension acts on particle B pulling it upwards.
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And since the string is light and inextensible, we know we don’t need to take into account its mass.
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The force acting downwards on body B is mass times acceleration due to gravity.
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So this time, that’s 𝑚 times 𝑔.
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Since the table is smooth, there are no other forces.
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If the table wasn’t smooth, we would need to consider friction.
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Finally, we know that when we release the system from rest, it will move as shown.
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Let’s call the acceleration with what we’re trying to find, 𝑎 centimeters per square second.
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At this stage, we might want to consider the forces on both particles.
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But actually, we have enough information just to use particle 𝑎.
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We’re going to recall Newton’s second law of motion.
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That is, the sum of the forces 𝐹 on an object is equal to the mass of that object multiplied by the acceleration 𝑎 of that object.
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The net force acting on particle A, in the horizontal direction at least, is 𝑇.
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Its mass is 590, and we’re looking for its acceleration; that’s 𝑎.
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Now, of course, we were told that the tension in the string is 90,860 dynes.
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So we replace 𝑇 with this number and we find that 90,860 equals 590𝑎.
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We solve for 𝑎 by dividing through by 590, such that 𝑎 is 90,860 divided by 590 which is equal to 154.
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And, of course, we’re working in centimeters per square second.
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So we’re finished.
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The acceleration in the system is 154 centimeters per square second.