Two Blocks, a Pulley and an Inclined Plane

In summary, the problem involves two masses connected by a string on an incline with an angle ##\theta## and an angle ##\alpha##. The goal is to find the value of acceleration by using the equations of motion. The magnitude of the acceleration is the same for both masses because the string does not stretch or shrink. The recommended approach is to orient the coordinate axes so that the acceleration is along one of the principal axes. This simplifies the equations and eliminates the need to introduce the angle ##\alpha##. It is important to determine which way the system will slide before starting the problem, as this determines the direction of the force of friction.
  • #36
Steve4Physics said:
Ignoring units: m1=55 and m2sinθ=199sin(39.3°)=126.
Since 126>55 we now know mass-2 moves downhill.
It took only a moment to find out!
But they are not in opposite direction. How can you subtract them?
 
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  • #37
rudransh verma said:
But they are not in opposite direction. How can you subtract them?
You do not need to subtract them. You just see which is biggest!

It's the same logic as described in Post #13 on your previous thread https://www.physicsforums.com/threads/two-bodies-hanging-from-pulley.1011640/
using the ‘Method 3’ approach.

Suppose you have two forces F1 and F2 applied to 2 (initially stationary) masses connected by a rope (represented as ======) all in straight line:

F1←[M1]=======[M2]→F2

The system accelerates left if F1>F2. (typo' corrected)
The system accelerates right if F2>F1. (typo' corrected)
(Assuming that |F1-F2| is sufficient to overcome any static limiting friction.)

Having the rope bent over a pulley doesn’t change this. The pulley changes the direction of the rope’s tension but not its magnitude. The pulley doesn’t affect in which sense acceleration occurs.

Edited -minor corrections.
 
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  • #38
rudransh verma said:
So am I correct?
You are not correct. You have
##\frac{F_n\sin⁡α−m_2g}{m_2}=-ay##
##\frac{m_1g}{m_1}=ay=g##
If you replace the value of ##ay## from the second equation into the first, you get

##\dfrac{F_n\sin⁡α−m_2g}{m_2}=-g##
##F_n\sin⁡α−m_2g=-m_2g##
##F_n\sin⁡α=0##.

Does this last result look correct? What is the equation for Newton's 2nd law for ##m_2## when the string is cut, i.e. when you have a single block sliding on an incline?

If you insist on using vertical and horizontal axes, you need keep ypurself honest and start with different accelerations for each mass.
Let ##\vec a_1## = acceleration vector of mass 1.
Under the assumption that the hanging mass is moving up, ##\vec a_1=a_1~\hat y##.

Let ##\vec a_2## = acceleration vector of mass 2. If the hanging mass is moving up, then ##m_2## is moving down and to the right (according to your FBD) ##\vec a_2=a_2\cos\!\theta~\hat x -a_2\sin\!\theta ~\hat y##.
Because the string is inextensible, you have to set the magnitudes equal, ##a_1=a_2=a##. The acceleration vectors for each mass are
##\vec a_1= a~\hat y~~## and ##~~\vec a_2=a\cos\!\theta~\hat x -a\sin\!\theta ~\hat y##.
Note that the accelerations of two masses have the same magnitudes but different y-components. As you did in post #1 you have to write three equations and make sure that you use the correct expressions for the acceleration components on the right-hand side. That is why I gave them to you.
 
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  • #39
Edit: These diagrams have been corrected as per post #42 below.
Block 1.png


Block 2 on slope.png
Block 1+2 on slope.png
 
Last edited:
  • #40
kuruman said:
Note that the accelerations of two masses have the same magnitudes but different y-components.
##-T\cos\theta+F_N\sin\theta=m_2a\cos\theta##
##T\sin\theta+F_N\cos\theta-m_2g=-m_2a\sin\theta##
##T-m_1g=m_1a##
After removing friction I got these. I think I have done it correctly. The acceleration magnitude is same but different y components.
##a=\sqrt{(a\cos\theta)^2+(-a\sin\theta)^2}##
 
  • #41
Steve4Physics said:
The system accelerates left if F1>F2. (typo' corrected)
The system accelerates right if F2>F1. (typo' corrected)
(Assuming that |F1-F2| is sufficient to overcome any static limiting friction.)
And I thought the process of finding the direction would be a lengthy process. You sounded like that.
 
  • #42
Lnewqban said:
https://www.physicsforums.com/attachments/296694

https://www.physicsforums.com/attachments/296692

https://www.physicsforums.com/attachments/296696
I don't understand the FBDs in post #39. It seems that there is an internal contradiction. The FBD for the hanging mass shows tension T = m1 g up and weight m1 g down which implies that the net force on m1 is zero which means that m1 has zero acceleration. The "straightened" out diagram for the two-mass system shows net horizontal force to be Fnet = (1.23 - 0.56 - 0.54) N = 0.13 N. The contradiction that I see is that m1 is not accelerating while the system as a whole is. It can be fixed by writing T = m1(a+g) in the first FBD and recalculating its value after solving for the acceleration. Also, in the straightened out FBD there is a typo: mass m1 should be 0.55 kg, not 0.099 kg.
 
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  • #43
kuruman said:
I don't understand the FBDs in post #39. ...
As usual, you are correct, professor.
I will try to correct the drawing tonight and edit that post.
Thank you!
 
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  • #44
@kuruman You didn’t answer. Am I right in post#40 ?
 
  • #45
rudransh verma said:
@kuruman You didn’t answer. Am I right in post#40 ?
Sorry, I didn't know you expected an answer from me. The equations in post #40 are correct but only if you are ignoring the friction between ##m_2## and the plane.
 
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