# Mass falling and pulling others on a rough surface

1. Sep 8, 2016

### Karol

1. The problem statement, all variables and given/known data
4 m masses, μ is the coefficient of friction. what is the tension and what should the maximum μ be to allow acceleration.

2. Relevant equations
Mass-acceleration: F=ma

3. The attempt at a solution
$$\left\{\begin{array}{l} mg-T=ma \\ T-3mg\mu=3ma \end{array}\right.~~\rightarrow~~a=\frac{1-\mu}{4}g$$
So, from here, μ≤1 in order for a to be positive. but logically:
$$3mg\mu\leq mg~~\rightarrow~~\mu\leq \frac{1}{3}$$

2. Sep 8, 2016

### Staff: Mentor

Check your working for solving simultaneous equations.

3. Sep 9, 2016

### rude man

You're mixing up static & dynamic friction coefficients. Your "logically" expression is correct if μ = μs.

4. Sep 9, 2016

### Karol

$$+\left\{\begin{array}{l} mg-T=ma \\ T-3mg\mu=3ma \end{array}\right.$$
$$mg-3mg\mu=4ma~~\rightarrow~~a=\frac{1-\mu}{4}g$$
The question didn't distinct between the coefficients, it mentions only one.

5. Sep 9, 2016

### HoodedFreak

I believe that should give you a = g(1 - 3μ)/4

Then you need to ensure that a is positive, g and 1/4 are positive constants so you have the restriction that: 1-3μ ≥ 0

6. Sep 9, 2016

### Karol

$$\left\{\begin{array}{l} mg-T=ma \\ T-3mg\mu=3ma \end{array}\right.~~\rightarrow~~a=\frac{1-3\mu}{4}g$$
Thanks HoodedFreak

7. Sep 9, 2016

### PeroK

The first part of the question asks you to find $T$. I assume this means $T$ in terms of $m, g$ and $\mu$. But, you've eliminated $T$ from your equations.

8. Sep 9, 2016

### Karol

That's simple:
$$T=(g-a)m=...=\frac{3+3\mu}{4}mg$$