# It's another block/incline question - plus friction and a force

It's another block/incline question -- plus friction and a force

This shouldn't be difficult. I've tinkered for a few hours, and I know I'm just missing some ticky thing about getting it setup right. I'm going to post the /whole/ question here, but really if somebody can just help me get started, I should be able to solve it.

In fact, I've already made an incorrect attempt...so the computer has given me new numbers. I'm truly more interested in learning the concept than getting the right answer...

## Homework Statement

Block M = 7.50 kg is initially moving up the incline and is increasing speed with a = .585 m/s^2.
The applied force F is horizontal.
The coefficients of friction between the block and incline are μs = 0.443 and μk = 0.312. The angle of the incline is 25.0 degrees.

Find:

(a) What is the magnitude of the force F?
(b) What is the normal force N between the block and incline?
(c) What is the magnitude of the force of friction on the block?

## Homework Equations

I know n should be mgcos(theta)
I know F=ma
and that the resistant force from the friction (And we'll use the kinetic friction, I believe) is friction*normal

## The Attempt at a Solution

I tried to start with finding "N" -- but I believe N is affected by the force being applied. I've tried breaking down the mass into x andy components, as well as the force, and the force from friction -- but since I'm not getting my normal force right, I'm not getting the resistant force from friction right, and that throws the whole thing off.

Where I'm getting super lost is this:
If the surface was frictionless, I could easily determine the force from the acceleration. Because friction is involved I need "N" -- but I can't figure out how to find "N" without "F" -- so lost as to how to find "F" from "a" without "N" to help me know what the friction-force is.

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kuruman
Homework Helper
Gold Member

Correct typoIf you don't know something give it a name and do some algebra. If you do things right, it will work out in the end.

Let N = the normal force and F = the horizontal force.

Can you do the following? (Never mind that you don't have a value for F or for N)

1. Find an expression for the sum of all the force components in a direction perpendicular to the plane.

2. Solve the above expression for N in terms of the other quantities.

3. Find an expression for the sum of all the force components in a direction parallel to the plane.

4. Set the sum of the "parallel to the plane" force components equal to mass times acceleration.

5. Solve this last equation for F.

I'm getting 2 equations and 2 unknowns. That's fine -- but solving for the unknown values does not give me even close to the right answers.

When solved as displayed below, I get 21 for F (Should be between 80 and 120) and 50 for N (Should be between 100 and 120).

kuruman
Homework Helper
Gold Member

The components of F are incorrect. For example, the component of F up the incline is Fcos25o, not F/sin25o. Likewise the component of F into the plane is not F/cos25o.

Ok, thanks for the help. I'm going to work on it right now -- but if that's wrong than my trig is screwy. Can we take a side trip and figure out what misconception I'm running on? Here's the way I'm visualizing it:

It seems (to me) like Fcos25 would give me F^2/X (Where X is the component of F pointing up the incline) -- where I'm trying to have F cancel out in the trig and leave me with X.

Have I put the θ in the wrong place? Let me submit this and stare some more...

Last edited:

Ok -- I had an aha moment, and realized that F is the hypotenuse, not one of the sides.

Let's take another crack at this.

kuruman
Homework Helper
Gold Member

Right. The magnitude of the vector is the hypotenuse here.

Last edited:

Still not there.

I know it's wrong because I'm given a range for N between 100 and 120, and I'm only getting 19. Again...not the foggiest idea what's wrong.

It's the direction of my F*sin (θ) arrow. That's increasing N, not decreasing. Going in for round 20.

kuruman