Double blocks on a incline plane

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big_bounce
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Homework Statement



See attachment (1)

what's forces act on pulley ?
Find direction and magnitude




Homework Equations


F=ma


The Attempt at a Solution


After we find tension in robe , See attachment (2)
what do we work ?
Must decomposes tension to horizontal and vertical vector and after that resultant of the two vectors ? or decomposes m1gsinθ and m2sinθ ?
See attachment (3) and then say is that Correct solution ؟
 
Attachments
  • Physics diagram of a triangular wedge with two frictionless-looking inclined faces at 45 degrees on the left and 60 degrees on the right, with a 5 kg block on the left face and a 3 kg block on the right face connected by a rope over a pulley at the apex; a red arrow labelled 'here' points to the pulley.
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  • Handwritten-style typeset equations solving two-block system: M1: W1 sin 45 - T = M1a (1), M2: T - W2 sin 60 = M2a (2), sum gives a = 1.14 m/s^2 and T = 28.94.
    2.PNG
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  • Force diagram of a pulley at the top of two inclines, with forces labelled 12 N, 13 N, 25 N, 22.50 N, 47.50 N, 50 N, 25.98 N and 35.35 N, and components of m1.g.sin(45°) and m2.g.sin(60°) drawn as vectors.
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All the pulley sees is the tension in the rope.
How many forces does the pulley see?
What is the magnitude of each of those forces?
Which direction does each force act?
What is the total applied force?
 
big_bounce said:

Homework Statement



See attachment (1)

what's forces act on pulley ?
Find direction and magnitude

Homework Equations


F=ma

The Attempt at a Solution


After we find tension in robe , See attachment (2)
what do we work ?
Must decomposes tension to horizontal and vertical vector and after that resultant of the two vectors ? or decomposes m1gsinθ and m2sinθ ?
See attachment (3) and then say is that Correct solution ؟
Please give a complete statement of the problem you're working on.

It seems that you are trying to do more than simply find the forces acting on the pulley.
 
you just add the two tensions vectorially and the answer u get is the correct answer... Tensions are the only forces acting on the pulley.m1gsin8 and m2gsin8 will act on the blocks, not on the pulley...
 
We want to calculate forces act on pulley .
So only tension's vectors act on pulley ?

See attachment
Is this correct answer ?
Resultant vector's magnitude and direction is correct ?


In this case F1=F2=T=28.94
 
Attachments
  • Vector diagram of forces F2 at 45 degrees and F1 at 30 degrees resolved into horizontal and vertical components, with resultant FR shown as the diagonal, labelled with the component expressions F2.sin(45°) - F1.sin(30°) and F2.cos(30°) + F1.cos(45°).
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The pulley is attached to the top of the triangular structure by bracket. This bracket exerts a reaction force upon the pulley.
 
lewando said:
The pulley is attached to the top of the triangular structure by bracket. This bracket exerts a reaction force upon the pulley.

And the reaction is equal to FR but in opposite direction , right ?

Please see Upper attachment
 
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Right! Numerically, what did you get for the magnitude and direction of FR?
 
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