# Newton's law to find acceleration

1. Feb 27, 2013

### Mangoes

1. The problem statement, all variables and given/known data

A block of mass m lies on a surface thirty degrees above the horizontal. A force with a magnitude equal to twice the weight of the block pushes it uphill. If the coefficient of kinetic friction is 0.4, find the acceleration of the block.

2. Relevant equations

Newton's laws of motion

3. The attempt at a solution

Drew free body diagram:

http://i.imgur.com/dA7MxvM.png

If I choose my coordinate axes so that the x-axis is parallel to the ramp,

ƩFy = 0
ƩFx = ma

First the x-components:

ƩFx = 2wsin60 - wsin60 - fk = ma

I know that fk = μn, I know μ but don't know n.

To find n from the y-components:

ƩFy = n - 2wcos60 - wcos30 = 0
n = w(2cos60 + cos30)

Substituting n into the x-component equation:

ƩFx = 2wsin60 - wsin60 - μ(w(2cos60 + cos30)) = ma

ƩFx = w(2sin60 - sin60 - 0.5cos60 + 0.25cos30) = ma

0.83w = ma

Weight is the product of mass and acceleration due to gravity

(0.83)mg = ma
0.83g = a

a = 8.134 m/s^2

This question was on a small quiz I took today and although I don't have the answer with me, I remember seeing that the answer was near 7.8 and not 8.134. Where am I going wrong with this?

2. Feb 27, 2013

### haruspex

you seem to be assuming the 2w force is horizontal. It is not entirely clear, but I would assume it's pointing up the slope.

3. Feb 27, 2013

### Mangoes

When I wrote the question in the original post I was just writing what I could recall from memory.

Now that it's been a while, I'm not so sure whether or not the real question stated a reason for believing the force was horizontal or in the direction of the slanted field... It's possible that I just goofed and assumed it was horizontal for some reason or I just can't remember but it did state it, but in any case, wouldn't the real answer have to be higher than the acceleration I provided in my post?

I guess a better question would be if my line of reasoning is correct (assuming the force is horizontal). While I don't remember the question word-for-word, I do know that it was the same reasoning I used in the quiz.

EDIT:

Nevermind, checked my work again and there was a sign error anyways. It might have been that I just goofed and assumed the force was horizontal for some strange reason...

The cos(30) is supposed to be negative instead of positive which leads to a much lower acceleration. The higher acceleration in my prof's answer might be due to what you said.

Last edited: Feb 27, 2013