Kinetics

WoAiMeiMei SG

I'm having trouble with this question. okay, we have 2 blocks one over each other. We have to draw a graph. Would the graph be (X-Y) ~~ (P - Friction), Friction force on Y and P on the x axis? if so , the graph would be like.. (Linear positive gradient line from zero, Hits a peak (μs.N), drops vertically down to (μk.N) and continues horizontally across the end. So how to figure out the gradient of the line and important critical points? FBDs?

WoAiMeiMei SG

Im guessing we can calculate the bottom block by using both masses as one combined mass (for N force), but wouldnt the acceleration be different for the top block when we calculate that one?

eagle SG

Hits a peak (μs.N), drops vertically down to (μk.N) and continues horizontally across the end.

Friction opposes motion... So it μs will give you the first P that acceleration starts (the vertical line).

Then right at the very start, the acceleration will be due to μk which you must calculate

you have to draw the free body diagram for each block and calculate the forces...

e.g. for Block B, you have friction backwards between B and ground, and friction backwards between A and B, and then P

for Block A, you have friction forwards between A and B (which will cause the forward acceleration of A).

So assuming no slipping, no sudden increase in P, we have for A
Max static friction = 0.20 * 36 * 9.81

For B, we have, at max static friction,
P - 0.20 * 36 * 9.81 - 0.15 * (0.36 + 0.45) * 9.81 = 0 (just start moving)
So we can calculate value of P where the acceleration goes vertically up.

Then to calculate the initial value of acceleration,
Use the above value of P

Plot P as x-axis, and acceleration as Y axis (function of P mah)

WoAiMeiMei SG

*posts up what i did" i feel something is missing because i didnt use "Pmax = 270N" but i think that was just a check.

WoAiMeiMei SG

i must be missing something..

eagle SG

Friction opposes motion... So it μs will give you the first P that acceleration starts (the vertical line).

Then right at the very start, the acceleration will be due to μk which you must calculate

you have to draw the free body diagram for each block and calculate the forces...

e.g. for Block B, you have friction backwards between B and ground, and friction backwards between A and B, and then P

for Block A, you have friction forwards between A and B (which will cause the forward acceleration of A).

So assuming no slipping, no sudden increase in P, we have for A
Max static friction = 0.20 * 36 * 9.81

For B, we have, at max static friction,
P - 0.20 * 36 * 9.81 - 0.15 * (36 + 45) * 9.81 = 0 (just start moving)
So we can calculate value of P where the acceleration goes vertically up.

Then to calculate the initial value of acceleration,
Use the above value of P

Plot P as x-axis, and acceleration as Y axis (function of P mah)

Made a small mistake in my posting in my rush just now before my tuition, not sure if you noticed... anyway...

Consider overall system...
P - 0.15 * (36 + 45) * 9.81 = 0 (just started moving)
So max P = 119.2 N

Once it starts moving, which is at P = 119.2 N,
kinetic friction between B and ground = 0.10 * (36 + 45) * 9.81 = 79.461 N

Hence, acceleration at this point = (119.2 - 79.461) / 81 = 0.49 m /s^2

Suppose no slipping, friction between A and B is the force pulling A forward.
So, frictional force = 0.49 * 36 = 17.64 N
Max static friction = 0.20 * 36 * 9.81 = 70.6 N

Since 17.64 N < 70.6 N, thus there is indeed no friction

The graph should be something like:

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The vertical rise is at P = 119.2 N, and the first value of acceleration = 0.49

And as P increases, it should be a straight line increase.

WoAiMeiMei SG

the difference between mine and yours is the second part,you calculated

"Hence, acceleration at this point = (119.2 - 79.461) / 81 = 0.49 m /s^2"

where i left mine in Newton units, and im still not to clear about the graph and how to draw it , (because on my graph my y-axis was Friction force). So we've calculated the values of Friction force , what about the x-axis for acceleration?

The graph that you drew would be different to mine, because mine would level out to a horizontally flat line. so im not sure about how to go about the graph now, and finding the critical points.

WoAiMeiMei SG

How do we take into account for when the a block slips off block b?

eagle SG

the difference between mine and yours is the second part,you calculated

"Hence, acceleration at this point = (119.2 - 79.461) / 81 = 0.49 m /s^2"

The graph that you drew would be different to mine, because mine would level out to a horizontally flat line. so im not sure about how to go about the graph now, and finding the critical points.

the graph would continue higher

afterall, F = ma, and with higher P, it will give you higher a, thus it cannot be a horizontal flat line

eagle SG

How do we take into account for when the a block slips off block b?

calculate the max kinetic friction, and divide by 36kg. That will give u the acceleration when the block starts to slip off B

From there, you can calculate the value of P

WoAiMeiMei SG

i mean, is there anything wrong with the graph i drew above (for block b and ground)?

It's a graph of Friction Force vs P, or wouldnt it be the same!?

eagle SG

If it is friction force vs P, you should be correct

WoAiMeiMei SG

what i still dont understand is the block A on block B graph :/

Im looking at yours and mine and im still not sure.

for your graph:

we know the vertical line is 119.2N

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However, I'm unsure about the linear gradient bit. how would i have to draw that, not sure of the angle etc..

eagle SG

dun calculate the angle... instead... calculate what happens as P increases... what happens to the acceleration of the whole thing (you are plotting acceleration vs P)

You have to visualise what happens... I think later after my tuition I try to do the whole thing and see if I can scan it...

Also, you have to understand what the question wants to ask... Plot acceleration of each block... that means before P reaches 270 N, the frictional force between A and B should exceed the static friction, which means A should eventually slip off... This is not confirmed... it's just from experience on how teachers would normally want to test students...

My tuition is starting soon.... so I try to do later... interesting question that I would love to try anyway :D

WoAiMeiMei SG

thanks... ill check back later

WoAiMeiMei SG

ugh, this acceleration business adds more complications

eagle SG

Repeat just now:

Consider overall system...
P - 0.15 * (36 + 45) * 9.81 = 0 (just started moving)
So max P = 119.2 N

Once it starts moving, which is at P = 119.2 N,
kinetic friction between B and ground = 0.10 * (36 + 45) * 9.81 = 79.461 N

As P increases...:
When A is just about to slip from A, static friction between A and B = 70.6N
so acceleration at this point = 70.6 / 36 = 1.9611 m/s^2
at this point, P = 1.9611 * 81 + 79.461 = 238.311 N (note: still smaller than 270N)

So as P continue to increase further, A will slip
Friction between A and B = 0.15 * 36 * 9.81 = 52.974 N (kinetic friction now)
At this point where P = 238.311 N,
acceleration of A = 52.974 / 36 = 1.4715 m/s^2 (means A slow down)
acceleration of B = (238.311 - 52.974 - 79.461)/45 = 2.353 m/s^2 (means B speed up)

And as P continue to increase further, and A continue to slip,
acceleration of A = 1.4715 m/s^2 because kinetic friction only...
acceleration of B will increase linearly...

And at P = 270, acceleration of B = (270 - 52.974 - 79.461)/45 = 3.057 m/s^2

Acceleration of A vs P

Acceleration of B vs P

WoAiMeiMei SG

Thanks For your help eagle, will have a look at your work. :D