# Incline Friction Problem?

## Homework Statement

A 2.00 kg block is held in equilibrium on an incline of angle = 75° by a horizontal force applied in the direction shown in Figure P4.50. If the coefficient of static friction between block and incline is µs = 0.300, determine the following.
(a) the minimum value of F
(b) the normal force exerted by the incline on the block

Illustration of Problem: http://www.webassign.net/sercp/p4-50.gif...

F=ma

## The Attempt at a Solution

For Y Component:
n-mgcos75=0
n=mgcos75
n=2.00*9.8cos75
n=5.07 N
It says this is wrong because of: Your answer differs from the correct answer by orders of magnitude.

For X
F-fs=0
F=usn=0
F=usn
F=0.300 x 5.07 N = 1.52 N

Does this look like I did it correctly?

Last edited:

learningphysics
Homework Helper
No. You have missing components. The F force has a component in the x direction (parallel to the incline) and y direction (perpendicular to the incline).

Also the mgsintheta component is missing (component of gravity parallel to the incline)...

And friction acts upward along the incline... it's preventing the block from sliding down. It works together with F to prevent it from sliding down.

Doc Al
Mentor
For Y Component:
n-mgcos75=0
n=mgcos75
n=2.00*9.8cos75
n=5.07 N
It says this is wrong because of: Your answer differs from the correct answer by orders of magnitude.
You neglected the Y component of the applied force F.

Ok, well now I really don't even almost have a clue what to do.

Of course the Force has x and y components, but if I don't know the force how would I solve to find those?

learningphysics
Homework Helper
Resolve F into its components using the angle.

Would the normal force be equal to mg?

That is, 19.6 N?

I am so confused, I don't even know what angle goes with F.

I can figure out the components of mg, but haven't a clue how to go about finding it for F.

learningphysics
Homework Helper
I am so confused, I don't even know what angle goes with F.

I can figure out the components of mg, but haven't a clue how to go about finding it for F.

It's the same theta. Just have to be careful about sin vs. cos.

From the tail of the vector F (the part without the arrow) draw a line perpendicular to the incline. And from where that line intersects the incline draw another line to the head of the vector F (the part with the arrow). See the two components? One of the angles of this triangle is theta.

learningphysics
Homework Helper
Would the normal force be equal to mg?

That is, 19.6 N?

I think I'm just screwed.

Thanks for the help anyways.

learningphysics
Homework Helper
I think I'm just screwed.

Thanks for the help anyways.

Draw an x - axis parallel to the incline... y -axis perpendicular to the incline... now rotate your picture so that the x-axis is horizontal, and y -axis is vertical... now try to resolve the components of F along the x-axis and y-axis.

So should the component parallel to N be pointing up (above F)?

learningphysics
Homework Helper
So should the component parallel to N be pointing up (above F)?

No sorry... it's opposite to N...

And fs should be upward along the plane... and the other component of F is parallel to fs upward along the plane.

Ok, so I drew it and there are 3 angles: 90, 45, and 45.

Which angle should I be working with?

Should I get Fsin45 for my x component and Fcos45 for my y-component?

Ok, I got it. It's n=35.64 N and F=31.66, correct?

learningphysics
Homework Helper

http://server4.pictiger.com/img/1317497/picture-hosting/force.jpg [Broken]

Fx = Fcos75
Fy = Fsin75

Last edited by a moderator:
learningphysics
Homework Helper
Should I get Fsin45 for my x component and Fcos45 for my y-component?

No. I was wrong it can't be any right triangle... look at the picture I uploaded...

learningphysics
Homework Helper
Ok, I got it. It's n=35.64 N and F=31.66, correct?

Yeah, I think that's right.

So should the component parallel to N be pointing up (above F)?

Here you will have to resolve two components. 1) mg 2) F.
For mg one of its componets[mgsin(theta)]is parallel to the plane which will be trying to bring the block downwards.Another would be perpendicular to the plane[mgcos(theta)](towards the plane).

For F one of its components would be perpendicular to the plane(towards the plane)[fsin(theta)] and another would be parallel to the plane which will be trying to pull the block upwards[fcos(theta)].

hence fcos(theta)-mgsin(theta)-mu{fsin(theta)+mgcos(theta)}=0
Now as it is on the verge of moving we consider a=o. so ma=o
Now you solve the above equation and you get the answer.

Sorry doc al, i had to give the equation.