# I'm stuck! Not sure where to go. Help!

1. Dec 5, 2007

### Cowtipper

1. The problem statement, all variables and given/known data
An object with a mass of 10 kg (this mass is m1) rests on a horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass of 4.0 kg (this mass is m2). The coefficient
of static friction between m1 and the horizontal surface
is 0.50, and the coefficient of kinetic friction is 0.30. If the system is set in motion with m2 moving
downward, what will be the acceleration of the system?
2. Relevant equations
m1 = mass one
m2 = mass two
T = tension
μ = coefficient of friction
g = gravity
a = acceleration
R = normal reaction

I got this helpful hint at another site, but I still can't figure it out...

For m2:
m2 g - T = m2 a ...(1)

Resolving horizontally and vertically for m1:
T - μR = m1 a ...(2)
R = m1 g ...(3)

Substitute for R from (3) in (2):
T - μm1 g = m1 a ...(4)

(m2 - μm1)g = (m1 + m2)a
a = (m2 - μm1)g / (m1 + m2)
= (4 - 0.3 * 10)g / (10 + 4)
= g / 14
= 0.701 m/s.

3. The attempt at a solution

And this is where I run into the problem. I'm not sure what to do, even with the helpful hint provided on another site.

Last edited: Dec 5, 2007
2. Dec 5, 2007

### hotcommodity

You initially stated that m1 rests on a frictionless surface, then you give coefficients of static and kinetic friction for m1 and the surface. How is this so?

3. Dec 5, 2007

### Cowtipper

Oops, that was my mistake. I've edited the problem - it isn't frictionless. I'm very sorry about that...

4. Dec 5, 2007

### hotcommodity

Don't sweat it, I just wasn't sure what the problem was trying to accomplish. I'm not sure what you're having trouble on, you've already arrived at the correct answer. The only problem is your units and sig figs. It should be .70 m/s^2.

5. Dec 5, 2007

### Cowtipper

Oh, okay. Thanks a for the assistance.

6. Dec 5, 2007

### vector3

$$\ W_2 - F_f = (m_1+ m_2) \cdot a$$ eq (1)

$$\ N = m_1 \cdot g$$ eq (2)

$$\ F_f = \mu \cdot N$$ eq (3)

$$\ F_f = \mu \cdot m_1 \cdot g$$ eq (4) from (2) into (3)

$$\ W_2 = m_2 \cdot g$$ eq (5)

$$\ m_2 \cdot g - \mu \cdot m_1 \cdot g = (m_1+ m_2) \cdot a$$

$$\frac{\((m2 - \mu \cdot m1)} {\((m2 +m1)} g = 0.701$$