# Equilibrium/friction problem

1. Sep 28, 2007

### mslena79

1. There is a block parallel and touching a vertical wall, a force F is applied at a 40 deg. angle from -x,-y. the weight of the block is 88.9N. The coefficient of static friction between the block and the wall is 0.560.
a) what is the minimum force F required to prevent the block from sliding down the wall?
b) what is the minimum force F required to start the block moving up the wall?

2. Relevant equations

3.a) Fs(max)=(0.560)(88.9N)=49.8N
Fy=88.9N-49.8N=39.1N
Fx=(39.1N)/(cos(40))=51.0N
$$\sqrt{39.1^2+51.0^2}$$=F
F=64.3N

b) 88.9N +49.8 N=138.7N=Fy
Fx=(138.7N)/(cos(40))=181.06N
$$\sqrt{138.7^2+181.06^2}$$=F
F=228N

2. Sep 28, 2007

You should be more specific...

I saw an error on your equations. The friction equation is defined as F$$_{sf}$$=$$\mu$$$$_{s}$$*N.

N is the normal force of the surface. In your case is the Normal force from the wall and you used the weight of the box. You should use as N the x component of the force F you should apply.

Try again knowing this fact.

Note: Superscript should be underscript.

3. Sep 28, 2007

### mslena79

a)88.9N*tan(40)=74.6N=Fn
Fy=88.9N
Fs=(0.560)*(74.6N)=41.8N
88.9N-41.8N=47.1N
$$\sqrt{47.1^2+74.6^2}$$=88.2N
F=88.2N

4. Sep 28, 2007

Are you guessing?

Here, the least force applied to prevent motion is when friction vector is pointing upward.
Write both of your equations, $$\Sigma$$$$\vec{F}$$=0 x and y.

Solve equations simultaneously. You unknowns are N and F and you have two equations.

5. Sep 28, 2007

### mslena79

I don't understand.

6. Sep 28, 2007

Due to my english? It can't be that bad.

7. Sep 28, 2007

### mslena79

No, I think I am having trouble because I am not sure where to place the applied force vector on my free body diagram. I have fs pointing +y, mg -y, Fn=-x, F 40 deg. in the (-x,-y)quadrant.

8. Sep 28, 2007

Ahh. That's they way it's suppose to be. Now you make use the following equations

$$\sum Fx=0$$

$$-Fn+F*cos(40)=0$$

Note: I assumed the angle given is the angle between the force and the x axis, otherwise you should use sin

$$\sum Fy=0$$

$$Fs-m*g+F*sin(40)=0$$ $$Fs=\mu s*Fn$$ so first equation for Fn you should have:

$$Fn=F*cos(40)$$ and you substitute this equation on the $$\sum Fy=0$$ equation:

$$F*cos(40)-m*g+F*sin(40)=0$$ and then $$F=\frac{m*g}{cos(40)+sin(40)}$$

;-)

Last edited: Sep 28, 2007
9. Oct 20, 2009

### jacksonpeeble

I recently came across this question while looking for solutions to mine (which was essentially the same thing). After solving it, I thought that I'd come back and add on to the solution, in case there were still some questions or somebody else is in need of assistance.

Fcos(x)-w+usFn=0 and Fsin(x)-Fn=0

After simplifying the two equations that, together, find the solution to the problem, one ends up with:

$$\frac{W}{cos(\Theta\degree)+\mu_{s}sin(\Theta\degree)}=F_{n}$$

where Theta is the angle measurement, Mu is the coefficient, and W is the weight (or force).

10. Oct 20, 2009

### jacksonpeeble

Likewise, for the second part, simply subtract the second term in the denominator rather than adding, because the frictional force is in the opposite direction.