# Calculating the frictional force

Here's my problem and the work I've done. The time is already past for me to submit the answer, but I want to know where I went wrong and why I was wrong.

The 2-kg box slides down a vertical wall while you push on it at a 45 degree angle from below. Both the box and the wall are wood. What magnitude of force should you apply to cause the box to slide down at a constant speed?

The coefficient of kinetic friction for wood-wood is 0.2.

The vertical forces acting on the box are:

$F_{box}$(sin 45) - $F_{friction} - mg$ = 0

where

$F_{box}$ = force acting on the box

$F_{friction}$ = frictional force opposing the motion

$m$ = mass of the box

$g$ = acceleration due to gravity

Hence

$F_{friction} = F_{box}$(sin 45) - $mg$ --- call this Equation 1

The normal force acting on the box is as follows:

$F_{box}$(cos 45) = $F_n$

and since

$F_{friction}$ = µ$F_n$, then

$F_{friction}$ = µ($F_{box}$)cos 45 --- call this Equation 2

Setting Equation 1 = Equation 2,

$F_{box}$(cos 45) - $mg$ = µ($F_{box}$)cos 45

Simplifying the above for "$F_{box}$"

$F_{box}$(cos 45) - $F_{box}$(µ*cos 45) = $mg$

$F_{box}$(sin 45 - µcos 45) = $mg$

and solving for "$F_{box}$"

$F_{box}$ = $\frac{mg}{cos 45 - µcos 45}$

Substituting appropriate values and calculating for "$F_{box}$"

$F_{box}$ = 34.65N

The system says that the solution is 23N. How did they get that and where is my mistake?

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## 1 Answer

The box is sliding vertically down at constant speed after overcoming your force F and friction, so the equation of forces in vertical direction should be

$mg = F_{box}(sin 45) + F_{friction}$

Repeat your calculation with the above equation, then you will get the correct answer.

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Damn, I looked at the problem and made the same mistake as Tyler :-) – John Rennie Oct 3 '13 at 7:01
thanks a lot. I wouldn't have seen that otherwise. I appreciate it. – Tyler Murphy Oct 4 '13 at 2:54