Inclined plane with friction and external force

[verhooverhoven]

Homework Statement

A block weighing 70.0 N rests on a plane inclined at 25.0° to the horizontal. A force F is applied to the object at 50.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.395 and 0.156.

Homework Equations

(a) What is the minimum value of F that will prevent the block from slipping down the plane?

(b) What is the minimum value of F that will start the block moving up the plane?

(c) What value of F will move the block up the plane with constant velocity?

The Attempt at a Solution

I calculate that the normal force is 63.441 N, which means the force of friction is 25.0594 N.

The force of the block sliding down the plane (without friction) is 29.5832 N, so the net force of the block sliding down is 4.5238 N.
The external force is at an angle of 50°, so the x component is F*cos(25).

4.5238 / cos(25) is 4.991, which is supposed to be the answer to A, but it is supposedly wrong. I have no idea what I am doing wrong.

 

[TSny]

Hello.

The normal force will be affected by the applied force F. So, the normal force is not going to be 63.4 N

Draw a good free-body diagram for the block and apply Newton's 2nd law for the net force parallel and perpendicular to the inclined plane.

 

[Simon Bridge]

In addition to TSny:

Homework Statement

A block weighing 70.0 N rests on a plane inclined at 25.0° to the horizontal. A force F is applied to the object at 50.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.395 and 0.156.

If A is the angle of the plane to the horizontal, then the angle of the force to the horizontal is 2A: what is the angle of the force to the plane?

(a) What is the minimum value of F that will prevent the block from slipping down the plane?

I calculate that the normal force is 63.441 N, which means the force of friction is 25.0594 N.

How? To get the best out of these forums, you should show your working.

The force of the block sliding down the plane (without friction) is 29.5832 N, so the net force of the block sliding down is 4.5238 N.
The external force is at an angle of 50°, so the x component is F*cos(25).

Which is the x direction - you have to explicitly state it as a definition.
If you mean the +x direction to point up the 25deg slope, then you need to check those angles.
Maybe by drawing a scale diagram and using a protractor - just to be sure.

Note: it is best practice to do the algebra first and substitute the numbers in at the end.
It makes things easier to troubleshoot and helps you see the relationships more clearly.