# Free body diagram for the static friction of an object with a horizontal force applied?

If an object is at rest on an incline, being pushed up in place by a horizontal force.

Does the component of the vector holding it up (relative to the incline), add to the normal force of the object, therefore increasing its static force of friction?

Does this mean that the static friction of an object is not a constant?

This is a sample question related to this idea that I am trying to understand if anyone is confused as to what I am asking:

A 5kg block rests on a 30 degree incline. The coefficient of static friction between the block and the incline is 0.2. How large of a horizontal force must push on the block if the block is to be on the verge of sliding up?

This is my free body diagram to try and understand this:

F_{\text{net}}=0\,\mathrm N\quad m=5\,\mathrm {kg}\\ \begin{align} F_F=0.2(42.4\,\mathrm N)&=8.48\,\mathrm N\\ 24.5+8.5&=33\,\mathrm N \end{align}\\[16pt] F_N=(5\times9.8)\cos(30)\\ \text{Static friction}=0.2(42.4)=8.5\,\mathrm N\\ 24.5+8.5=33\,\mathrm N\\[12pt] \frac x{\sin90}=\frac{33}{\sin60}\quad x=38\,\mathrm N\\[16pt] \text{Answer is}\,43\,\mathrm N

• @user104372 The frictional force is down the slope (opposite to the direction of potential movement) and does not have a magnitude of 8.5 N. The normal reaction is made larger by the component of the horizontal pushing force which is perpendicular to the slope. – Farcher Sep 22 '16 at 10:26
• @user104372 Just suppose the frictional force was large enough to hold the mass on the slope without the external force. The frictional force would be up the slope. Now apply a gradually increasing horizontal force until such a value that no frictional force was need to keep the mass in position. Now increase the horizontal force and the frictional force will now be acting down the slope to prevent the mass moving up the slope. – Farcher Sep 22 '16 at 11:23
• Sorry for the late reply, I was in classes until now. Using the information here I was able to brute force an answer by trying to skirt around the value of friction. I don't feel satisfied with it but when I wake up I'll re-read all these posts to see if I can conceptually understand what is going on. This is my work to the solution: imgur.com/a/64ylq – user130666 Sep 23 '16 at 7:07
• @Scibeon I have posted a free body diagram for you. – Farcher Sep 23 '16 at 8:21
• According to the answer key in the textbook I am reading, 43N is the answer for the horizontally applied force. Unless I somehow got that answer through error. – user130666 Sep 23 '16 at 19:19

You have four forces acting on the body.
A horizontal force, a frictional force, a normal reaction and the weight of the body.
Your free body digram does not show all four forces.

You need to apply Newton's second law parallel to the plane and perpendicular to the plane and the perpendicular application should answer your question about the normal force and hence the frictional force.

Update

Here is my free body diagram and yours with the normal reaction and the frictional force added.
Note the $F \sin 30^\circ$ component of $F$ which is pushing down on the object and thus making $N$ greater than $mg \cos 30^\circ$ by that amount and so increasing the value of the frictional force along the slope.

Using Newton's second law you can two equations with two unknowns, $F$ and $N$, and hence solve for $F$.

• Is this diagram including all the forces? imgur.com/a/pc8Vg – user130666 Sep 22 '16 at 6:50
• I cannot talk you through the problem step by step. In you diagram I count five forces. As I wrote before I think that there are four forces acting on the mass. Draw the forces, give them symbols and apply Newton's second law. You are also still insisting that the normal reaction is equal to the component of weight perpendicular to the plane. It is not as you have missed the fact the horizontal force has a component in that direction. – Farcher Sep 22 '16 at 7:06
• I believe that I understand the free body diagram now. N points up, and the applied force adds to the force of friction. Thank you very much for correcting my diagram and to everyone else that gave their input. – user130666 Sep 23 '16 at 19:18

Force due to static friction is coefficient of friction * Normal reaction. As the body is resting on an inclined plane, the normal reaction is not equivalent to weight, but the component of it perpendicular to the plane on which the body rests. You need to resolve all the forces in your diagram, to arrive on your answer.

From what I see, you are going good equations are concerned, but be a bit descriptive when it comes to the diagram, to arrive on the correct conclusion.

Like downward force is mgsin30 already acting on the block. Normal reaction is the cos30 of weight, so you find the friction which acts in direction opposite to the motion and you know the rest.

• Is this diagram including all the forces? imgur.com/a/pc8Vg – user130666 Sep 22 '16 at 6:50