# Laws of motion

a block of mass 4.8 kg is pushed up against a wall by a force F that makes 37 angle with the horizontal. the coefficients of static and kinetic friction are 0.2 and 0.1 respectively. the magnitude of the force F in N that makes the block just about to slip down:

wut i did is:
F sin 37-0.1+mg=0
Fcos37=n

but i didnt get the answer right which is 63.2 !!!

## Answers and Replies

Doc Al
Mentor
F sin 37-0.1+mg=0
Redo this correctly: The sum of forces in the vertical direction must add to zero.
Fcos37=n
Good.

Is the given angle above or below the horizontal?

i think its above horizontal
but even when i add them up the answer is wrong
and by solving Fcos37=n the answer of F is wrong too !!

Doc Al
Mentor
Please redo your first equation. (Until you fix the first equation, don't expect to be able to solve for F correctly.)

What are the vertical forces? Which way do they act?

the first equation Fsin37+0.1+mg=0
the vertical forces acts upward because its abt to slip down
right ?

HallsofIvy
Science Advisor
Homework Helper
Doc Al, I notice mooneh is using "0.1", the coefficient of kinetic friction, in his equations. Since the mass is "just about to slip down" but not moving, shouldn't that be coefficient of static friction?

Doc Al
Mentor
the first equation Fsin37+0.1+mg=0
the vertical forces acts upward because its abt to slip down
right ?
Gravity acts upward? What's 0.1 supposed to represent? What's the friction force? (Is it static or kinetic?)

Doc Al, I notice mooneh is using "0.1", the coefficient of kinetic friction, in his equations. Since the mass is "just about to slip down" but not moving, shouldn't that be coefficient of static friction?
Yep. But that's not the only problem. here is wut i did:
Fsin37+0.2-mg=0
Fcos 37 =n
F ( cos37+sin37) = 93.88
F = 67

Well you fixed one thing, broke another, and are missing a big thing in both cases >_>

There are three forces acting on it, friction, the guy pushing, and gravity. Gravity and friction are both trying to pull it down, the guy pushing is trying to push it up. Since they're in the same direction, gravity and friction need to be both negative or both positive, and then the guy pushing should be the opposite

As of your previous post you have...actually I'm looking at your work and maybe you did it more right than it seemed

You typed Fsin(37)+.2-mg=0, and I'm looking at the .2, .2 by itself is not the force of friction, you must multiply by the normal force. But then you found the normal force and used it but I dunno where the .2 went or the 93.88 came from. You have the right idea though

Remember, force of friction = coefficient * normal force

Fgrav+Ffr-F*sin37=0, or F*sin37-Fgrav-Ffr=0, whichever

Last edited:
Doc Al
Mentor
here is wut i did:
Fsin37+0.2-mg=0
0.2 is just the coefficient of static friction; the friction force is $f_s \leq \mu N$. Since we assume the block is just about to slip, the friction force must have its maximum value (in terms of the normal force).

Fsin37-0.2n-n=0
Fsin37= 1.2(Fcos37)
???

Doc Al
Mentor
Fsin37-0.2n-n=0
Which way does the friction force act? What happened to the weight?

blochwave said Fgrav+Ffr-F*sin37=0, or F*sin37-Fgrav-Ffr=0
thats why i changed my answer and put negative signs to both Ffr and weight
weight=mg=n

Doc Al
Mentor
blochwave said Fgrav+Ffr-F*sin37=0, or F*sin37-Fgrav-Ffr=0
thats why i changed my answer and put negative signs to both Ffr and weight
But since the block is just about to slide down, friction acts up while gravity acts down; they have different signs.
weight=mg=n
Weight = mg, but does not equal the normal force!

The above is just the equation for vertical forces. Combine this (after you fix it) with your equation for the horizontal forces (from your first post).

Fcos37=n
Fsin37+ 0.2n - mg=0

5 Fsin37 + n -235.2 = 0
Fcos 37 - n =0

F(5sin37+cos37)=235.2
F=61.7

Doc Al
Mentor
Looks good to me! but the correct answer is 63.2 !!
never mind close enough :)
lool u must hate me now
thx alooooooooot
ur the best