A-level Maths Mechanics: Prevent Block from Sliding Down Inclined Plane

[nab_]

Homework Statement

A block rests on a slope which is angled at θ° to the horizontal. The coefficient of friction between the surface of the slope and the block is tan α. P1 is the horizontal force that needs to be applied to the block to stop it from slipping down the slope, whilst P2 is the greatest horizontal force that can be applied without the block slipping up the slope. Show that P2/P1= (tan(θ + α))/(tan(θ - α))

Relevant Equations

F= μR

The best I could do was draw a forces diagram. I know that friction would be working up when the block is on the point of slipping down the plane and friction will be acting down the slope against the direction of motion when the block is on the point of slipping up the slope. (not even sure if I have that right at this point)

I don’t know how to go about this problem at all.

 

[haruspex]

I know that friction would be working up when the block is on the point of slipping down the plane

So post your diagram for that case and see what equations you can write.

 

[nab_]

So post your diagram for that case and see what equations you can write.

does that look right

 

[PeroK]

View attachment 299742
does that look right

You need to do this in two parts. First you need to analyse the case where we have force $P_1$ and the block is almost slipping down. Then, when we have the force $P_2$.

Note that your normal force depends on $P_1$ and $P_2$.

This is hard for A-level! You'd have been better off with media studies!

 

[nab_]

You need to do this in two parts. First you need to analyse the case where we have force $P_1$ and the block is almost slipping down. Then, when we have the force $P_2$.

Note that your normal force depends on $P_1$ and $P_2$.

This is hard for A-level! You'd have been better off with media studies!

I don’t understand what would happen to the normal force as it goes from P1 to P2. I thought the only thing that would change is the direction of friction.

It is hard. I didn’t come across any similar problems in practice questions but I think we have covered all the necessary stuff to apply to the question.

 

[PeroK]

I don’t understand what would happen to the normal force as it goes from P1 to P2. I thought the only thing that would change is the direction of friction.

What if $\theta = 90$ degrees? What's the normal force then?

 

[nab_]

What if $\theta = 90$ degrees? What's the normal force then?

Equal to P1 ?

 

[PeroK]

Equal to P1 ?

Exactly. The normal force depends on $W, P$ and $\theta$.

 

[nab_]

Exactly. The normal force depends on $W, P$ and $\theta$.

Oh ok so because P is acting horizontally into the plane the normal reaction force would increase. So also add in -Psin θ when calculating for the resultant forces perpendicular to the plane?

 

[PeroK]

Oh ok so because P is acting horizontally into the plane the normal reaction force would increase. So also add in -Psin θ when calculating for the resultant forces perpendicular to the plane?

I'm not sure why you have a minus sign there.

The first thing is to write down the equation for $R$. Can you do that?

 

[PeroK]

... and then the (balanced forces) equations for $P_1$ and $P_2$.

 

[nab_]

I'm not sure why you have a minus sign there.

The first thing is to write down the equation for $R$. Can you do that?

is this right so far

 

[berkeman]

is this right so far

Just a quick side note -- please learn to post math equations using the "LaTeX Guide" link below the Edit window. That makes your work *much* more legible than uploading dark pictures of your hand-written work. Thanks.

 

[PeroK]

Okay, to give you some Latex to copy and work with we have:
$$R = W\cos \theta + P\sin\theta$$I must admit, the next equation I would have written down is:
$$F = \mu R = \mu(W\cos \theta + P\sin\theta)$$And you have:
$$W\sin \theta - \mu R -P_1 \cos \theta = 0$$Note that at this point the general $P$ becomes $P_1$.

Now, you need to work on that a bit to isolate and get an expression for $P_1$.

 

[nab_]

View attachment 299748
is this right so far

sorry I had the signs muddled and also wrote tan (theta) instead of alpha.

so I rewrote the equation to get
$Wsin \theta -tan \alpha Wcos \theta -tan \alpha Psin \theta -Pcos \theta$ =0

And then rearrange for P:
P= $\frac {Wsin \theta -tan \alpha Wcos \theta }{tan \alpha sin \theta +cos \theta }$

this would be the equation for P1 and then I’d I write the equation for P2. after that I should be able to simplify P2/P1 to get the equation stated in the question right?

 

[PeroK]

Looks right so far.

 

[nab_]

Looks right so far.

I still wasn’t really able to get the right answer. I don’t know where I’m going wrong but I keep hitting a dead end.

thank you so much for your help :)

 

[berkeman]

I still wasn’t really able to get the right answer. I don’t know where I’m going wrong but I keep hitting a dead end.

So post your work; that's what we are here for.

And thanks for picking up the basics of LaTeX -- that is a huge improvement!

 

[haruspex]

Looks right so far.

Not to me, but I don't have time right now to pin down the error. Looks like a sign is wrong in post #12.
Edit: in view of post #15, I don’t think there is enough of the working that arrives at the equation there to say what the error is. But I strongly suspect a sign error, making two factors that ought to be different look the same, thereby erroneously cancelling them.
Hint: the difference in the equations for the two cases should be the same as negating the friction coefficient.

 

[PeroK]

I still wasn’t really able to get the right answer. I don’t know where I’m going wrong but I keep hitting a dead end.

thank you so much for your help :)

I would also work on the final expression with the tangents to express that in terms of $\sin \theta, \cos \theta$ and $\mu$, using the appropriate trig identities. Then check that this expression matches $P_2/P_1$.

 

[PeroK]

Not to me, but I don't have time right now to pin down the error. Looks like a sign is wrong in post #12.
Edit: in view of post #15,

The equation in post #15 is correct:

$P_1 = \frac {Wsin \theta -tan \alpha Wcos \theta }{tan \alpha sin \theta +cos \theta }$

 

[haruspex]

The equation in post #15 is correct:

You are right - I must have misread it.
So I'd like to see what the OP did next.

 

[nab_]

You are right - I must have misread it.
So I'd like to see what the OP did next.

So I got these two equations for P2
$W \sin \theta +F - P \cos \theta = 0$

$R=P \sin \theta +W \cos \theta$

together:

$W \sin \theta + \tan \alpha (P \sin \theta + W \cos \theta ) - P \cos \theta =0$

and then rearranged;

$P = \frac{-\tan\alpha W\cos \theta -W\sin \theta }{\tan\alpha\sin\theta-\cos\theta}$

I don't know if things are wrong here or if they start going wrong when I cross multiply the two fractions and simplify.

 

[PeroK]

I'm not sure how you got that minus sign. And why you're writing $P$ instead of $P_2$.

Sort that out and write down the ratio

 

[nab_]

I'm not sure how you got that minus sign. And why you're writing $P$ instead of $P_2$.

Sort that out and write down the ratio

$P_2 =\frac{\tan\alpha W\cos\theta + W\sin\theta}{\cos\theta-\tan\alpha\sin\theta}$

what do you mean by ratio

 

[nab_]

I'm not sure how you got that minus sign. And why you're writing $P$ instead of $P_2$.

Sort that out and write down the ratio

OH I GOT IT!
so I managed to simplify both $P_1$ and $P_2$ after factoring out $\cos\theta$

$P_2= \frac{\tan\alpha+\tan\theta}{1-\tan\theta\tan\alpha}=\tan(\theta+\alpha)$
$P_1=\frac{\tan\theta-\tan\alpha}{\tan\theta\tan\alpha+1}=\tan(\theta-\alpha)$
which by the sum and difference rule would give me the correct answer

 

[PeroK]

OH I GOT IT!
so I managed to simplify both $P_1$ and $P_2$ after factoring out $\cos\theta$

$P_2= \frac{\tan\alpha+\tan\theta}{1-\tan\theta\tan\alpha}=\tan(\theta+\alpha)$
$P_1=\frac{\tan\theta-\tan\alpha}{\tan\theta\tan\alpha+1}=\tan(\theta-\alpha)$
which by the sum and difference rule would give me the correct answer

There is a factor of $W$ missing, but otherwise that's it.

The bonus question is what contraints are there on $\mu$ and $\theta$ for these formulas to hold?

E.g. if $\mu$ is large enough, then $P_1=0$.

 

[nab_]

There is a factor of $W$ missing, but otherwise that's it.

The bonus question is what contraints are there on $\mu$ and $\theta$ for these formulas to hold?

E.g. if $\mu$ is large enough, then $P_1=0$.

I don't know. can $P_1=0$ even if the plane is not completely horizontal ( $\theta=0$)

 

[PeroK]

I don't know. can $P_1=0$ even if the plane is not completely horizontal ( $\theta=0$)

Yes, for a low angle, it's likely that the block will not slide (unless there is not much friction).

And, for a steep angle and enough friction there will be no solution for $P_2$.

This is worth noting since it was not immediately obvious that $P_1$ and $P_2$ would be valid for the same scenario. I actually checked that as part of my solution.

In any case, for any $\theta$ there is an upper bound on $\mu$ for this problem.

 

[nab_]

In any case, for any $\theta$ there is an upper bound on $\mu$ for this problem.

ahh okay. I think I understand that

 

[PeroK]

ahh okay. I think I understand that

This sort of thing is definitely worth noting. We have two force equations:

$P_2= \frac{\tan\alpha+\tan\theta}{1-\tan\theta\tan\alpha}$
$P_1=\frac{\tan\theta-\tan\alpha}{\tan\theta\tan\alpha+1}$

The equation for $P_1$ is only valid when $\tan \theta > \tan \alpha = \mu$ and the equation for $P_2$ is only valid when $1 - \tan \theta \tan \alpha > 0$. I.e $\mu < \cot \theta$.

That gives the overall condition that $\mu < \tan \theta, \cot \theta$.