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Normal forces on inclined plane

  • #1
A 1.0 kg box on a 30 degree frictionless incline is connected to a 3.0 kg box on a horizontal frictionless
surface. The pulley is frictionless and massless.

I got n=mg for m1 in the y direction, and in the x direction i got F+T=m1a. For m2 in the y direction i get N-m2gcos(theta)=0, and in the x direction i get m2gsin(theta)-T=m2a. Is this correct?
 

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Answers and Replies

  • #2
Ba
101
0
Start with the box on an inclined plane, the gravity pulls down and the slope pushes perpendicularly to itself, then the normal force for the second box on the flat plane is mg the tension actually makes no differance to the normal force. If you are looking for all the forces then the tension is the horizontal component of the first normal force
 
  • #3
691
1
You should have two FBD's for this problem. It makes things a lot easier IMO. Draw A fbw for both boxes as if they are on a flat surface. Draw one box with a tension force to the right (top box) and the other with a tension force to the left (bottom box). Here's the tricky part--the weight of the inclined box does not act normal to the surface of the plane (normal means 90 degrees btw or perpendicular). How many degrees is the weight of the box from the normal direction of the inclined box? Next, if I told you the weigh of the inclined box was the hypotenuse of a triangle would that help you resolve the weight vector into x-y vectors (hint hint).

good luck.
 

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