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Resolution of components

  • Thread starter logearav
  • Start date
  • #1
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Homework Statement



Revered Members,
Kindly see my attachments.

Homework Equations



This is Atwood machine dealing with motion on a frictionless inclined plane.

The Attempt at a Solution


To better visualize the forces involved, i isolated m1 and m2 and drawn free body diagrams for both. For mass m1, there are three concurrent forces, that is, T,m1g,and N, where T is the force in the string because of Tension and N is the normal force of the table on the block. The assumption is m1 accelerates up the plane, which is taken in x direction. Now, the weight m1g is broken down into components. The x component is in the assumed direction of acceleration , and the y component acts perpendicularly to the plane and is balanced by the normal force N.
I have shown the components in the attachment1. My question is
1) why m1gcosθ and m1gsinθ can't be interchanged, as given in my second attachment?
2) what is wrong in changing the angle θ, that lies between m1g and m1gcosθ( first attachment) to between m1gsinθ and m1g?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

Answers and Replies

  • #2
993
13
You can either work with θ or with (90 - θ), the latter angle you can call β. But, of course, you cannot call them both θ in the same problem.
 
  • #3
338
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But invariably in all the books i referred , the Tension is always balanced by the mgcosθ and not by mgsinθ. My teacher said, that 90 - θ should be the sine component always and θ should be cos component. Is that so?
 
  • #4
993
13
See the attachment.
 

Attachments

  • #5
338
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Oh! Beautifully explained grzz. Thanks a lot. I got it now.
 

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