Need help with a m1 friction question

I got the right answer by using your hint so thank you for that. But it doesn't make sense though because friction can be the same in both diagram and surely it is not right to assume that friction is the same in both cases....

Hey, wondering what the answer was as I've worked through it for practice so am curious if I got it right :P
Glad you got it sorted!

No worries! Yeah, that's right but unfortunately it is assumed that they are the same because otherwise you cannot solve the question.

At first, i thought you have to assume u is the same for both diagram...

The angle of the slope is greater than that at limiting friction so the frictional force is at its limiting value - it is also moving which means that the downward force of gravity has overcome the friction.

The frictional force acts of course up the slope - it opposes [or in this case tries to oppose] motion.

If you post your working I'll check it for you

Well, $\mu$ is definitely the same as it's a constant calculated for that slop and specific object. It's the reaction force that is not constant in both case.

I did post my work and I do understand limiting equilibrium. I think it means that when it is 30 degrees, the object is not moving because friction is equal to the other force going in the opposite direction which is why it stays still. Whereas for the second one :45 degrees, the other force is greater than the max friction and thats why it is accelerating downwards. What i don't get in this question it is why friction stays constant in both diagram although R is different. If it is based on assumption, then why not assume u is the same for both diagrams and go from there? But u can't be the same though because of different reaction forces. Does this question actually make sense?? ThANKS

so it is pointless to work out u then?

You couldn't work it out as you don't know the mass of the object.

I left m as mass and they cancel out at the end. But I got 2.9 by using u

Looking at it again, yeah, you can work out the coefficient of friction, but I got 0.29m/s² for the acceleration of the body. How did you work that out?

Edit: yes, I got 2.93m/s² (3 s.f) as well as I hadn't taken g into my calculation.

I somehow managed to but thats not right, so don't worry. Thanks for helping

Yeah, and I highly doubt the question required you to take account of the change in friction but here is my method:

When the object is in equilibrium, you can work out the horizontal components to the plane:
Let's take $\theta=30$ and the angle between reaction force and the horizontal component $\alpha=60$. Therefore,

$F_1 \cos \theta=R_1\cos\alpha$,

$F_1=\dfrac{R_1\cos\alpha}{\cos \theta}$ , but $R_1=mg\cos\theta$

$F_1=\dfrac{mg\cos\theta\cos\alpha}{\cos \theta}=mg\cos\alpha$

Therefore:
$\mu=\dfrac{F_1}{R_1}=\dfrac{mg \cos\alpha}{mg\cos\theta}=\dfrac{ \cos\alpha}{\cos\theta}$

For when the angle of inclination is increased, that is when the raction force has decreased, we have:

$R_2=mg\cos45$

$mg\sin45 -F_2=ma$
But $F_2=\mu R_2=mg\cos45 (\dfrac{ \cos\alpha}{\cos\theta})$

So $mg\sin45 -mg\cos45 (\dfrac{ \cos\alpha}{\cos\theta})=ma$

$\rightarrow g(\sin45 -\cos45 (\dfrac{ \cos\alpha}{\cos\theta}))=a$

I hope this is correct.

I got similar as that as well but I think the question is poorly worded because friction can be the same in both. Thank you

No worries! I just posted my method to show why I thought the friction is not constant in both case but yeah, that wasn't supposed to be a substitute for the answer.

Hi, sorry to bother but could you possibly give me some help on this question http://www.thestudentroom.co.uk/showthread.php?t=3794861&p=61558317#post61558317 thanks