# Why do I have to make gravity positive to make this work?

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1. Oct 3, 2015

### Shivang Saxena

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

Two objects are connected by a light string that passes over a frictionless pulley as shown in the figure below. m1 = 7.10 kg, m2 = 7.10 kg, and ϕ = 56°. When released from rest, m1 accelerates downward at 0.972 m/s2. For this to happen, the coefficient of kinetic friction must be ______ , and to even begin sliding in the first place the coefficient of static friction must be _____ (less/greater) than _______

I set it up so my up and right is positive

m1 = m2 = 7.10kg
a = -0.972 m/s

Θ = 56

2. Relevant equations

For mass 1,

∑Fy = T - mg = ma
T = m (a + g)

(No forces in x direction)

For mass 2,

∑Fy = N - mgsinΘ = ma = 0
N = mgsinΘ

∑Fx = -T + Ff + mgcosΘ = ma

Ff = μ * N
Ff = μ * mgsinΘ

3. The attempt at a solution

I have a working solution, I'm just not satisfied with why it works. The equations above are all I needed, and after some substitution and simplification I got:

μk = [ m ( -a - g) + m(gcosΘ -a) ] / mgsinΘ

The reason I'm not satisfied with this solution is that when I solve it, it only works if I plug in a as (-0.972), which it specified, but for gravity I have to put in (9.8). And even then I get -0.29 when it should be positive.

Why can't I put negative gravity in for g if my upward direction is positive? I searched around and found that the pos/neg depends on how you set up your sum of forces, but I have both positive and negative gravity in there and I'm very confused.

Also, I'm not sure how to approach the static friction part of the question. Shouldn't it also be 0.29? If someone could give me a hint as to where to start that would be great!

Thanks

2. Oct 3, 2015