Power screw horizontal applicaion FBD

1. Aug 3, 2010

M.E. Tom

Hello,

I have been trying to set up a free body diagram for a power screw in a horizontal application.
Every example I find they are using a power screw to lift or lower a load not move it left to right.

I have attached the example from the text book, along with my work so far.

Looking at the text book example why do they break the Normal force in to components when summing forces? In most physics text books the Normal force is left intact and other forces, such at weight, are broken down.

Second, on the work2.pdf when I am summing forces for the left traverse condition, I get the Force of friction in the x equal to the negative x component of the normal force. Is this right?

Third, should there be a thrust force causing the nut to move horizontally?

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2. Aug 4, 2010

Skrambles

Both methods are valid. It doesn't matter where your axis is located as long as you keep using it throughout the problem. If weight was the only force acting in that example then you would need to use less trig by placing one axis along the inclined plane. That way you wouldn't need to break the normal and friction forces into components.

Your answer is right, but it only works when the screw is not moving. If it was moving you would need set F=ma instead of F=0. Remember that the coefficient of friction is different when it is moving than when it is static.

3. Aug 10, 2010

M.E. Tom

Ok, this makes perfect sense. I drew the FBD not thinking about it moving and then confused myself when summing forces.

When it is moving then "P" will have to provide both the Torque and the Acceleration force. How would you draw that resultant force?