# Laws of motion

1. Jan 11, 2008

### mooneh

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 !!!

2. Jan 11, 2008

### Staff: Mentor

Redo this correctly: The sum of forces in the vertical direction must add to zero.
Good.

Is the given angle above or below the horizontal?

3. Jan 11, 2008

### mooneh

i think its above horizontal
and by solving Fcos37=n the answer of F is wrong too !!

4. Jan 11, 2008

### Staff: 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?

5. Jan 11, 2008

### mooneh

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

6. Jan 11, 2008

### HallsofIvy

Staff Emeritus
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?

7. Jan 11, 2008

### Staff: Mentor

Gravity acts upward? What's 0.1 supposed to represent? What's the friction force? (Is it static or kinetic?)

Yep. But that's not the only problem.

8. Jan 11, 2008

### mooneh

here is wut i did:
Fsin37+0.2-mg=0
Fcos 37 =n
F ( cos37+sin37) = 93.88
F = 67

9. Jan 11, 2008

### blochwave

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: Jan 11, 2008
10. Jan 11, 2008

### Staff: Mentor

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).

11. Jan 11, 2008

### mooneh

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

12. Jan 11, 2008

### Staff: Mentor

Which way does the friction force act? What happened to the weight?

13. Jan 11, 2008

### mooneh

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

14. Jan 11, 2008

### Staff: Mentor

But since the block is just about to slide down, friction acts up while gravity acts down; they have different signs.
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).

15. Jan 11, 2008

### mooneh

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

16. Jan 11, 2008

### Staff: Mentor

Looks good to me!

17. Jan 11, 2008

### mooneh

but the correct answer is 63.2 !!
never mind close enough :)
lool u must hate me now
thx alooooooooot
ur the best