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Course: Physics archive>Unit 3

Lesson 7: Treating systems

Two masses hanging from a pulley

In this video David explains how to find the acceleration of two masses hanging from a pulley (using the easy method). Created by David SantoPietro.

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• Is the tension for the 2 boxes is the same?
• Yes. The tension in the rope is equal and upwards on both the masses. This is how he is able to cancel the forces out in the video.
• How did he get 36.75 for tension? He used his calculator, but I wanted to see how he solved it because I got a different answer.
• a=F/m,
-2.45(m/s^2)={T-(gravity)}/5,
-2.45*5=T-mg,
T=-12.25+5*9.8=-12.25+49=36.75.

Therefore, T=36.75(N)

*T=tension, a=accelaration, F=force, m=mass, g=gravitational accelaration(g is approximately 9.8),

a=F/m came from newton's second law.

T-Gravity=F, because, T is towards + direction, and Gravity is towards the opposite direction(-)
• What would happen to the acceleration if the mass and radius of the pulley were taken into account? Such as if mass(pulley)=5kg and radius was 0.25m?
• If you assume the pulley has a non-negligible mass, then you have to factor in what is known as the "moment of inertia" of the pulley. The moment of inertia (symbol I (uppercase i)) is the rotational equivalent of regular inertia to motion. The formula for the m.o.i. of a pulley is 1/2mr^2, where m is the mass and r is the radius. So the m.o.i. of your pulley would be I=1/2*5kg*.25m^2=.156kg*m^2. The product of the m.o.i. and the angular velocity is going to equal the torque (rotational force) on the pulley. The tension in the rope, in turn, is equal to the torque on the pulley multiplied by the radius r. It's late and my answer is getting really drawn out but I only just learned about moment of inertia and its corresponding complications to these types of problems in my last semester, in my third year of university. There's a good sample problem here: http://farside.ph.utexas.edu/teaching/301/lectures/node112.html . For your reference the dot over a symbol means its derivative (this is calculus if you don't know); v dot is equal to accleration, and omega dot (looks like w, represents angular velocity) is equal to angular acceleration. I have no idea what year of school you're in, but as you can maybe see just taking the mass of a pulley into account makes everything significantly harder!
• , i miss one step how did he got 36.75 N what did he do with the aceleration?
• accelaration of the system is the same as accelaration of the object.
a=-2.45(m/s^2)

*accelaration is - because it is in - direction(downwards.)
(1 vote)
• how did you solve for t on the end part? ive been trying to rearrange it but i keep getting a different number
• Kind of random, but, if you are free falling and your acceleration due to gravity is 9.8 m/s^2, and your velocity increases by 9.8 m/s because of this, but your acceleration stays the same. Why is it that we feel more force when we hit the ground from higher up, when it is only our velocity that changes, and not our acceleration?
• You speed increases by 9.8 m/s every second, that is why it is m/s^2. So after the first second you moving at 9.8 m/s and after the second you are going 19.6, 29.4 after 3 seconds, 39.2 m/s after 4 second and 49 m/s after 5 seconds. This is why you hit harder when you fall from further up.
• How would you solve this the "hard way"
• Two free body diagrams, two sets of equations of motions, solved simultaneously.
• Is the Tensions on 3kg and 5kg still the same if we apply a friction on the rope?
• No, because the 5kg block is heaver then the 3kg block, so the 3kg block would go up and the 5kg block will go down so it won't be the same.
• are the tensions the same for both boxes, 3kg and 5kg
(1 vote)
• Absolutely the same. if you use the newton's second law in vertical direction of the 3kg mass you have to use the acceleration in the positive direction because it moves upward. so the acc = +2.45m/s^2 and then plug in the force of gravity of 3kg mass to find T and you will find that it has the same force as you find it on the 5kg mass by using acc = -2.45m/s^2.

I hope that help. :)
• if the pulley has a mass, when drawing the free body diagram will it have any other forces or just mg?