How Do You Calculate Friction and Normal Force on an Incline?

[Naeem]

Q. A Man, is pushing a copier machine, up an incline, which weighs 600 N.

I need to find the frictional force, and the normal force.

The free body diagram, I shall describe that to you.

Weight of the machine acting downards. and if we resolve this weight we get mg sin theta along the incline , and mg cos theta acting downward.

Friction is acting in the direction opposite to the applied force. The force applied is 180 N, at an angle of 10 degrees, which when resolved, given F cos theta in the x and F sin theta in the y.

Normal force is acting perpendicular to the surface.

Now, if we were to sum the forces in the x and y

x : F cos theta - F friction - mg sin theta = 0
y : F sin theta + Normal force - mg cos theta = 0

I have the theta, for the F cos theta which is 10 degrees, and the weight of the block which is 600 N.

How do I find, out the angle which the machine makes, with the incline, so that this may be used, to calculate mg sin theta & mg cos theta.

I've spent nearly 2hours with this problem, and have seemed to exhausted all possibilities. Guys, any ideas please help !

 

[Gokul43201]

Problems with your presentation :

1. Post the question EXACTLY as it appears in the text/source. Then post your attempt below that. There should be no confusion about what is stated in the question, and right now, there is.

2. If the angles of the incline and applied force are different, call them by different names (don't call them both "theta"). If the applied force acts "at 10 degrees", is that 10 degress above the horizontal, up the incline, or 10 degrees above the slope of the incline (or something else) ?

 

[Naeem]

The applied force is 180 N, at 10 degrees above the slope of the incline.

 

[Gokul43201]

Okay, I previously misunderstood what you were saying - I apologize.

There does not appear to be sufficient data in the question to find all the unknowns.

Please post the EXACT question.

 

[Naeem]

Believe Me Gokul,

I shall type the entire question, from the book, and I shall describe the figure from the book to the best of my ability.

Q. A man is pushing a copy machine that weighs 600 N up an incline. A free-body diagram if the man is shown in E4.6.38

a. If the net force in direction t is zero, what is the value of F friction ?
b. If the net force in direction n is zero, what is the value of F normal ?
c. What is the value of the ratio F friction / Fnormal ? (Note: The friction coefficient between the man;s shoes and the incline must be at least as great as this ratio if his feet do not slip as he walks up the incline.)

Now I shall describe the free-body diagram (Exactly what is drawn below the question in my textbook)

There is man, on an an incline, with a arrow pointing on him downward (600N) , the man has both his hands stretched pushing (another arrow at an angle 10 degrees pointing at his hands Force 180 N). Below the slope there is Friction pointing along the slope in the directon he is pushing, and one more arrow indicating F normal indicating upwards.

The directions t and n are in the from of a tilted co-ordinate axes. t along the incline, and n perpendicular to it.

Also, above the figure, to the top right there is an x and y coordinate axes, just showing which direction is positive (Here to the right is positive x , and up is positive y )

I hope this helps all of you who are reading this post, in determing that 'Theta'

 

[Naeem]

And in case you wanted to know what the answers were,

F normal = 503.5 N
F friction = 318.4 N
ratio = 0.632

 

[Gokul43201]

It is not clear whether you are describing forces on the copy machine or forces on the man. Clearly, the 600N force acts only on the copy machine, but the friction on the copy machine must act down the slope, not up it.

In any case, there's still insufficient info. The question is wrong.

 

[Naeem]

Yes, I agree with what you are saying. The 600 N , force is actually the weight of the copying machine but, The arrow is pointing downward on the man in the picture. Also, the frictional force must act in the opposite, direction, but I described the picture "AS IS" to you from the book. Do you think, we still don't have enough info, to find the angle "theta"...
May be there is an error. the figure, is missing the angle or something. That is what I think...

I think I should leave this problem for discussion tommorow, with the professor in class or with the TA...

Anyway thanks for your time.

 

[Gokul43201]

I think I should leave this problem for discussion tommorow, with the professor in class or with the TA...

Yes, do this.