# Finding max acceleration with force at angle incl friction

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1. Apr 18, 2015

### YearnToLearn

1. The problem statement, all variables and given/known data
Hey. I was doing some exam practice questions, but I hit a snag with this one and can't quite work out how to proceed.

A 6kg block at rest is pulled along a horizontal surface by force F→ at angle θ. Given that the coefficient of kinetic friction is 0.15, find the optimal angle at which to apply the force to achieve maximum acceleration.

2. Relevant equations
F = ma
fk = μkN
F→2 = F2sin2θ + F2cos2θ

3. The attempt at a solution
N = mg - Fsinθ
μk = 0.15/58.86
fk = (2.55×10-3)(mg-Fsinθ)
Fnetx = Fcosθ - fk
ax = (Fcosθ - fk)/m

From here I guess I need to form a differential equation and solve for maximum but this leads to θ being a ridiculous angle. Any advice? (P.S, sorry if this is in the wrong section, still trying to gauge the levels of physics being done in each)

2. Apr 18, 2015

### Delta²

You just have to take the derivative of $a_x$ with respect to $\theta$ and set it to 0 and solve the equation involving $\theta$, that is to solve $\frac{da_x}{d\theta}=0$.

3. Apr 19, 2015

### YearnToLearn

Still getting a weird value for θ while solving for the maximum. Will work through and check my values, but can we confirm that the logic is sound?

4. Apr 19, 2015

### Delta²

Do you get $tan(\theta)=\mu_k$ at the end? Why is this weird, since the friction coeeficient is small we expect theta to be small also (if friction coefficient was zero it would be theta=0 as can be understood easily).

There is something i dont understand why you divide 0.15 / 58.86 for $\mu_k$?

5. Apr 19, 2015

### YearnToLearn

My bad. Playing catch up at the moment and only started looking at this concept today. Another look at the derivative and I can see the manipulation. Thanks for the help