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Issy04
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Homework Statement


Solve for the forces P and N acting on an object in equilibrium on an inclined plane.
Weight=10N, angle is 30 degrees
http://blob:https%3A//www.physicsforums.com/d431a152-5681-411b-ba57-809320539165
Free-body diagram of a block on a 30 degree incline with forces N normal to the slope, P horizontal to the right, and a downward weight labelled 10.

Homework Equations


Trigonometry, i.e. sin, cos, tan rules. The 'Z' rule (amongst others) to determine angles.

The Attempt at a Solution


I have attempted the problem (as done with similar problems, which were correct) however the answers seem to assume that the angle in solving P is 30.
I have used in the Parallel P=10tan60=17.32N. However the answer is given by 10tan30=5.77N. Is the answer wrong? Or have I made inaccurate assumptions?
Hand-drawn free-body diagram of an object on a 30 degree incline, with forces N (normal, perpendicular to the slope), P (horizontal), and a downward 10 weight, with red 60 degree angle markings.

In the perpendicular, N=10sin60=8.66N, however this is wrong also! The answer is given by 10/sin60
Free-body diagram of an object on a 30 degree incline with forces N (normal, perpendicular to the slope), P (horizontal) and a 10 N weight acting downward, with red 60 degree angles marked at the object and the lower corner.

Thankyou for any help.
 
Assume a frictionless plane.

Resolve N into X and Y directions and it will become clear.
 
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insightful said:
Assume a frictionless plane.

Resolve N into X and Y directions and it will become clear.
The Y component of N is used to support 10N object from moving down vertically. Since we resolve N into components, then there must also be an X component.
Once you resolved N into components, N does not exist anymore but 4 forces acting on the object.
 
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azizlwl said:
The Y component of N is used to support 10N object from moving down vertically. Since we resolve N into components, then there must also be an X component.
Once you resolved N into components, N does not exist anymore but 4 forces acting on the object.

Thank you. I have calculated the correct value of P now but I am still confused as to why N does not equal 10sin60, as P has no vertical component?
 
The resultant of the 10 N force and P must be equal and opposite to N.

Free-body diagram of a particle at equilibrium on a 30 degree incline: normal reaction N perpendicular to the slope, a 10 N weight vertically down, horizontal force P to the right, and a red arrow labelled -N opposite N, with 30 and 60 degree angles marked.
 
Attachments
  • Free-body diagram of a particle on a 30 degree incline with forces N normal to the slope, P horizontal, and 10 downward, with the reversed resultant -N drawn in red.
    upload_2015-11-22_6-51-55.png
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Issy04 said:
Thank you. I have calculated the correct value of P now but I am still confused as to why N does not equal 10sin60, as P has no vertical component?
Now we try to solve using triangular method since net forces equal to zero.
As you you know you can place the vector, the arrow anywhere, parallel and pointing to same direction.
10N is the base, P is opposite, and N is hypothenus.
 
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azizlwl said:
Now we try to solve using triangular method since net forces equal to zero.
As you you know you can place the vector, the arrow anywhere, parallel and pointing to same direction.
10N is the base, P is opposite, and N is hypothenus.
Thank you very much! It is clear now.
ehild said:
The resultant of the 10 N force and P must be equal and opposite to N.

View attachment 92220