Mass falling and pulling others on a rough surface

[Karol]

Homework Statement

4 m masses, μ is the coefficient of friction. what is the tension and what should the maximum μ be to allow acceleration.

Homework Equations

Mass-acceleration: F=ma

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:
$$3mg\mu\leq mg~~\rightarrow~~\mu\leq \frac{1}{3}$$

 

[NascentOxygen]

Check your working for solving simultaneous equations.

 

[rude man]

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

 

[Karol]

Check your working for solving simultaneous equations.

I add the equations:
$$+\left\{\begin{array}{l} mg-T=ma \\ T-3mg\mu=3ma \end{array}\right.$$
$$mg-3mg\mu=4ma~~\rightarrow~~a=\frac{1-\mu}{4}g$$

You're mixing up static & dynamic friction coefficients

The question didn't distinct between the coefficients, it mentions only one.

 

[HoodedFreak]

I add the equations:
$$+\left\{\begin{array}{l} mg-T=ma \\ T-3mg\mu=3ma \end{array}\right.$$
$$mg-3mg\mu=4ma~~\rightarrow~~a=\frac{1-\mu}{4}g$$

The question didn't distinct between the coefficients, it mentions only one.

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

 

[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

 

[PeroK]

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

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.

 

[Karol]

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

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