Dynamics ramp and friction. Finding angle?

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britt6

Homework Statement


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.

Homework Equations


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

The Attempt at a Solution


m=3.0kg
l=2.0m
a=1.6m/s^2
ff=10Nif 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
 
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britt6 said:
ma-ff/mg=sintheta
What are you plugging in for ff here? This may be where the mistake is.
 
NFuller said:
What are you plugging in for ff here? This may be where the mistake is.

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.
 
britt6 said:
Therefore I plug in Ff kinetic to complete that idea.
Oh, right. I missed that they gave you the force of friction in the problem.
britt6 said:
ff+fgparallel=ma
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.
 
britt6 said:
mu?
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NFuller said:
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.
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).