0
$\begingroup$

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

enter image description here

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?

$\endgroup$
3
  • $\begingroup$ When you substituted 1) into 4), you may have dropped a minus sign. Otherwise, good job! $\endgroup$
    – LouisB
    Nov 16, 2015 at 6:43
  • $\begingroup$ @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. $\endgroup$
    – Jack
    Nov 16, 2015 at 14:14
  • 1
    $\begingroup$ 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" $\endgroup$
    – pentane
    Nov 16, 2015 at 19:45

1 Answer 1

0
$\begingroup$

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 $$\mu = \frac{T\sin{21^\circ}}{mg - T\cos{21^\circ}}= \frac{8.02(0.358)}{9.8-8.02(0.934)}=1.24$$

There is a minus sign in the denominator. You must have added instead of subtracting, to arrive at 0.166.

I use the cross product for the torque because I tend to make too many mistakes the other way. It would help if I made a good sketch (free body diagram) of all the force components. Without one, I need to use the cross product.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.