# Block and Pulley Problem with Friction [closed]

1.) What is the acceleration of the hanging mass? (Indicate the direction with the sign of your answer.)

2.) Determine the tension in the cord. (Indicate the direction with the sign of your answer.)

Seems like the simplest problem but I cant seem to get the right answer. Here's my approach:

$n$ = normal force on $m_1$

$f_s$ = friction force on $m_1$

$T$ = tension in the cord

$\mu_k$ = coefficient of kinetic friction

Problem 1.)

$(1)\quad n = m_1 g = (40kg)(9.8m/s^2)$

$(2)\quad n = 392 N$

$(3)\quad f_s = \mu_k n = (0.22)(392N)$

$(4)\quad f_s = 86.24N$

$(5)\quad T = m_2 g - f_s$

$(6)\quad T = (13.4kg)(9.8m/s^2) - 86.24N$

$(7)\quad T = 131.32N - 86.24N = 45.08N$

$(8)\quad m_2 a_{m_2} = m_2 g - T$

$(9)\quad (13.4kg) a_{m_2} = 131.32N - 45.08N$

$(10)\quad (13.4kg) a_{m_2} = 86.24N$

$(11)\quad a_{m_2} = 86.24N/13.4kg = 6.44m/s^2$

$6.44m/s^2$ is wrong, though. Could someone point me in the right direction? Any feedback is welcome, and thanks in advance

SOLUTION:

Equation (5) should be $T-f_s = m_1 a$

$T-f_s = m_1 a$

becomes (after rearranging)

$T-(40kg)a = 86.24N$

and

$m_2 a = m_2 g - T$

becomes (after rearranging)

$T + (13.4kg)a = 131.32N$.

Then we throw those two equations into a matrix:

\begin{bmatrix} 1 & -40 & 86.24 \\ 1 & 13.4 & 131.32 \end{bmatrix}

Solve, and our answers will be:

$a = 0.84m/s^2$

$T = 120.01N$

## closed as off-topic by jinawee, Brandon Enright, Kyle Kanos, user10851, DavePhDJun 8 '14 at 22:24

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• In equation (5) I changed $m_1$ to $m_2$ and carried out the rest of the problem with the same equations, and ended up with $-13.02m/s^2$, which is still incorrect. Is there anything else with my work I should change? – Garmenarnar Jun 8 '14 at 17:16