# Dynamics ramp and friction. Finding angle?

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1. Jul 26, 2017

### britt6

1. The problem statement, all variables and given/known data
A 3.0 kg block is released from the top of a rough ramp of length 2.0m and accelerates down the ramp at 1.6m/s^2. If a force of kinetic friction of 10 N acts on the block, what is the angle of the ramp.

2. Relevant equations
Fnet=ma
ff=(mu??? I don't have that symbol HAHA) muFn

3. The attempt at a solution
m=3.0kg
l=2.0m
a=1.6m/s^2
ff=10N

if Fnet=ma..... then

ff+fgparallel=ma
ma-ff=fgparallel
ma-ff=mgsintheta
ma-ff/mg=sintheta
then take inverse to get theta..

Now what! answer should be 30 but when I put in the values I get 10 degrees. help please

2. Jul 26, 2017

### NFuller

What are you plugging in for ff here? This may be where the mistake is.

3. Jul 26, 2017

### britt6

I just thought because the question states that there is a kinetic force of friction and acceleration, there are unbalanced forces, meaning you need an Fnet. Therefore I plug in Ff kinetic to complete that idea.

4. Jul 26, 2017

### NFuller

Oh, right. I missed that they gave you the force of friction in the problem.
You have a sign mistake here, it should be
$$-F_{f}+F_{g}=ma$$
because the vectors for friction and gravity point in opposite directions.

5. Jul 26, 2017

### haruspex

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That depends on the sign convention chosen. If Britt chooses the same direction as positive for all forces and accelerations (a good approach often) then the equation for parallel to the plane is $$\Sigma F=F_{f}+F_{g}=ma$$
The negative sign comes in when values are plugged in. If downslope is positive then Ff=-10N, Fg=+mg sin(θ).