# Finding the coefficient of friciton [closed]

I'm asked to find the coefficient of friction before the bar starts sliding.

Here is my solution ->

∑M = 0

-mg * 1.4863 + 1.8159*T = 0

(T*1.8159) i found by using a cross product ( -2.9726i+ 2.6765j ) cross (-Tsin21i +T cos21j)

∑Fx = 0

-Tsin21 + Ffr = 0

∑Fy = 0

-mg + N + Tcos 21 = 0

1) from ∑M = 0, T = 8.02m

2) from ∑Fx = 0, -Tsin21 + uN = 0

3) from ∑Fy = 0, N = mg - Tcos21

Substitute 3) to 2)

4) -Tsin21 + umg - uTcos21 = 0

Substitute 1) to 4)

8.02sin21m + umg + u*7.487m = 0

from here i get u = 0.166

But the correct answer is 1.24. I have checked it over 10 times, but everything seems correct. Please, can someone tell me where i made a mistake?

• When you substituted 1) into 4), you may have dropped a minus sign. Otherwise, good job! Nov 16, 2015 at 6:43
• @Louis but that will not impact the final answer. I still get the wrong answer even though i checked it more than 10 times.It is not recommended to use cross product for finding torque? Because i tried doing it without cross product and i got the correct answer.
– Jack
Nov 16, 2015 at 14:14
• Hi Jack, check out this meta post about what kind of questions are on topic here. "It's not enough to just show your work and ask where you went wrong. If you just need someone to check your work, you can always seek out a friend, classmate, or teacher. As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on" Nov 16, 2015 at 19:45

In your equation (1), $T = 8.02 m$ and in your equation (4) $-T\sin{21^{\circ}} + \mu mg - \mu T \cos{21^{\circ}} = 0$. Rearranging (4) and then substituting the value of T from (1) gives $$\mu = \frac{T\sin{21^\circ}}{mg - T\cos{21^\circ}}= \frac{8.02(0.358)}{9.8-8.02(0.934)}=1.24$$