Calculating Tension in a Pulley System at an Angle

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To calculate tension in a pulley system at an angle, it's essential to draw free-body diagrams for each block involved. The slope's orientation and the direction of the pulling force relative to the surface are critical factors. Providing a diagram of the setup can clarify the forces at play, including the tension acting on block B2 opposite the pulling force. It's important to ensure that the weight components in the equations are accurately represented. Understanding these dynamics will help in solving for the tension effectively.
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Homework Statement
A 1.7kg book B2 is connected by a rope and pulley to a 6.3kg book B1.
We neglect friction.
B2 is on a horizontal surface, and B1 is on a 36.1º slope.
If the force pulling B1 is 28.2N what is the tension between the two books?
Relevant Equations
T=gm
Hi!
I really can't figure this one out...
I have a = (F-cos(36.1)g) = a and from that I get T = mB1 a = 6.3 (20.3736) = 128 N.
Could someone please help?
Thanks!
 
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You need two equations that you get by drawing two free-body diagrams, one for each block. Is the slope above or below the horizontal? Is the pulling force parallel to the surface on which B2 slides? Please post a diagram.
 
Last edited:
Where is the pulley? Please show a picture of the complete setup as was given to you.
 
Screen Shot 2021-09-28 at 7.06.33 PM.png
 
Thank you for the drawing. There is tension T acting on B2 in a direction opposite to F, no? Ideal pulleys change the direction of the force but not its magnitude. Also, check the components of the weight in your equation for B2.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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