Minimizing Pulling Force for Crate on Rough Floor

[ethex]

Homework Statement

A person is dragging a packing crate of mass 100kg across a rough floor where the coefficient of kinetic friction is 0.4. He exerts a force F just sufficient to keep the crate moving at a constant velocity. At what angle above the horizontal should his pulling force F be for it to be minimum?

Homework Equations

The Attempt at a Solution

Kinetic friction force = 0.4 x 100g = 40g

Fcos theta = 40g
F sin theta = 100g

tan theta = 10/4
theta = 68.1 degree

am i right?

 

[grzz]

Draw a diagram and put in ALL the forces. You are leaving out the normal reaction force of floor on crate. This normal reaction force IS NOT EQUAL to the weight of the crate since the pulling force is not horizontal.

 

[ethex]

Ok.

Let the angle be theta,

Friction force = 0.4 x 100g x cos theta

F - 100g sin theta - 40g cos theta = 0

F = 980 sin theta + 392 cos theta

dF/dtheta = 980 cos theta - 392 sin theta

Let derivative = 0

tan theta = 2/5

i got back the same ans! What is wrong?

 

[gneill]

Ok.

Let the angle be theta,

Friction force = 0.4 x 100g x cos theta

Why? What does cos($\theta$) have to do with the weight of the crate?

What's the normal force?

 

[grzz]

The suggestion was to DRAW A DIAGRAM WITH ALL THE FORCES. Then you can get more help.

 

[grzz]

The suggestion was to DRAW A DIAGRAM WITH ALL THE FORCES. Then you can get more help.