Friction On 2 Blocks + Angled Pull

[sandman203]

Homework Statement

Hi all, greetings from aus! having mad trouble with this question:

so I've figured out that the friction limit for block B is 11.35N when it is stationary. Putting my co-ords parrallel to the 15 degree incline means the force due to gravity in the x direction is 5.0728N.. so effectively ill need to make up the 6.2772N so the block B will slide down block A.

Now I've toyed with the idea of makign F Large enough to lift the whole body to an angle where the force due to gravity will overcome friction. However there are no dimensions on the block so i can't make an equation for work needed. SO block A will remain horizontal and slide across. I just don't understand how force F will be able to make block B fall down if it doesn't angle up? It;s probably got something to do with Fy but i just can't see it. Any help would be greatly appreciated. Have been racking my brain for like 5 hrs obver this.

 

[TSny]

Hello, sandman203. Welcome to PF!

If F causes the system to accelerate, then the normal force between A and B will be reduced and so the friction force between A and B will be reduced. If the acceleration is enough, B will begin to slip.

The best approach to these types of problems is the standard recipe of drawing free-body diagrams and applying Newton's laws.

See what you can deduce from Newton's 2nd law applied to a free-body diagram of block B when block B is just on the verge of slipping.

 

[verty]

Hint: what is happening at the interface between B and A? What is making B accelerate?

 

[haruspex]

I just don't understand how force F will be able to make block B fall down if it doesn't angle up? It;s probably got something to do with Fy but i just can't see it.

More Fx than Fy. Imagine pulling the wedge left with a sharp yank. What do you think would happen to the block?

 

[sandman203]

i imagine the block would slide right off if the acceleration is high enough.. I just don;t know how to set that up mathematically. Do i want to break down F in terms of the co-ord system i used on block B, so fx is parrallel to the 15 degree incline? then i will will a force going up towards block B (reducing the normal force?). I don't see how fx acts on block B, does it do anything at all? obviously up until it surpasses the friction limit. am i on the right track here?

 

[TSny]

How many forces act on block B? Can you describe them?

 

[sandman203]

We have gravity going straight down. This is broken into 2 forces Fxb (Parrallel to the incline) and Fxy (perpandicular) to the incline + the normal force Nab which is equal to Fxy. We also have a force of friction opposite to Fxb which is helping the block stay stationary atm. When F is applied then i get confused. Does all components of F get applied to block B? I am thinking maybe only an amount until it reaches the friction limit. If block A is pulled to the left will there be an opposite reaction from block B? causing acceleration in the opposite direction??

 

[TSny]

We have gravity going straight down. This is broken into 2 forces Fxb (Parrallel to the incline) and Fxy (perpandicular) to the incline + the normal force Nab which is equal to Fxy. We also have a force of friction opposite to Fxb which is helping the block stay stationary atm. When F is applied then i get confused. Does all components of F get applied to block B? I am thinking maybe only an amount until it reaches the friction limit. If block A is pulled to the left will there be an opposite reaction from block B? causing acceleration in the opposite direction??

The force F is applied to block A. It does not act on block B. So, you have 3 forces acting on B:

(1) Force of gravity
(2) Normal force (from the surface of A and perpendicular to the surface)
(3) Friction force (from the surface of A and parallel to the surface)

Draw a diagram showing just block B and these 3 forces acting on B.

Next, you want to introduce a coordinate system for block B. It is generally a good idea to choose one of your axes of your coordinate system to be in the direction of the acceleration of the block. As long as B is not yet slipping on A, think about the direction of the acceleration of B. Choose your x-axis in that direction and then the y-axis will be perpendicular to that.

Finally, set up Newton's 2nd law for block B for the x and y components of motion:

$\sum$F_x = m_Ba_x
$\sum$F_y = m_Ba_y

Then see what you can deduce from these equations.

 

[sandman203]

yeah i have drawn the free body diagram many times.. i don't know what the hell I am missing here dude. I not seeing how force F reduces the normal force and hence friction ?

 

[haruspex]

yeah i have drawn the free body diagram many times.. i don't know what the hell I am missing here dude. I not seeing how force F reduces the normal force and hence friction ?

While everything is stationary, the block experiences a normal force, a frictional force acting up the plane, and gravity. These balance both horizontally and vertically. If the wedge now accelerates to the left and the block stays with it, there must be a net horizontal force on the block to provide that acceleration. This changes the equations. Pls post those equations.

 

[sandman203]

This is where I am at:
Block B

X dimension is parrallel to incline, y dimension is perpandicular

When stationary

ƩF_y = m_ba_y = 2 × - 9.8 + N_ab = 0

ƩF_x = m_ba_x = 19.6 sin 15 - F_f = 0

N_ab = 2 × 9.8

Limiting friction force for this surface = 0.6 × N_ab = 11.76N

Block A When stationary

* Do i put co-ords parrallel to the ground here? so the 2 blocks have different co - ords? or stick with the same co - ords i used for block B?

ƩF_y = m_ab × a_y = 7 × - 9.8 + N_a = 0 so N_a = 68.6

Limiting friction for for this surface = 0.6 × N_a = 41.16N

ƩF_x = m_ab × a_x = 0

When F is applied

ƩF_y = m_ab × a_x = - F cos30 + F_f

So we can see that in order to make AB move we need a_x = (41.16 - F cos30)/7

I don't know where to go from here.

 

[haruspex]

You want to know the max F such that B moves with A. So suppose B does move with A, but the friction between them is at maximum. If the acceleration is a, what are the force equations for B (I suggest using horizontal and vertical resolution)? What are the equations for the combined system? You should have four equations and four unknowns: normal force between A and B, acceleration, frictional force from the ground, and F.

 

[sandman203]

Cool dude, what about friction force between AB isn't that another unknown?? as block B accelerates with block A it will have x/y components, this will adjust the normal and hence friction force right?

 

[sandman203]

Also, when F is applied does it only take into account block B or do the equations just need mass of A... i guess if we apply enough to override friction then B's mass won't matter, is this correct?

 

[sandman203]

answer!

Hey guys, thanks for the help i have managed to come up with an answer! so here we go.

First of all here are my fbd's of block A and block B

Now for the math:
When stationary - Block A

ƩF_x = m_Aa_A = m_A × 0 = 0

ƩF_y = m_Aa_A = 0 = -49N + N_A

∴ N_A = 49N

When Stationary - Block B

ƩF_x = m_B × g × sin15 - F_fab = 0

ƩF_y = - m_B × g × cos15 + N_AB = 0

∴ N_AB = m_B × g × cos15

When F is applied - Block A

ƩF_x = m_A × -a_x = F cos30 + F_fa

ƩF_y = m_A × a_y = 0 = N_a - 49 + F sin30

∴ N_a = 49 - F sin 30

∴ Static Friction Limit → F_fA = 0.6 × (49 - F sin30)
When F is applied - Block B

ƩF_x = -2a_x cos15 + (m_B × g × sin15) ± F_fAB

ƩF_y = 0 = N_AB - (m_b × g × cos15) - 2a_x sin15

∴ N_AB = (m_b × g × cos15) - 2a_x sin15

∴ Static friction limit → F_fAB = 0.6 × 2×g×cos15 + 2a_xsin15

SO we want to find the acceleration in which B needs to go in order to make equilibrium (anymore N and block B will slide down)

∴ F_x = 0 = -2a_xcos15 + (2×g×sin15) - 0.6 × 2×g×cos15 + 2a_xsin15

Solving for a_x = - 4.445m/s²

So Now we find out how this acceleration works on Block A.

We will need ƩF_x = -4.445 x 5 = -22.225N in the x direction from force F to act on Block A and put block B in equilibrium.

F_x = -22.225N - F_fA

F_x = -22.225N - (0.6×49 - Fsin30)

Now F_x = Fcos30 = -22.225 - 0.6(×49 - Fsin30)

Solving for F we get 44.5N!

So this will keep block B on block A so i guess the answer is F > 44.5N ?

 

[verty]

You have some mistakes and I think you are still not quite knowing what to do.

Start again but from this point of view: A and B are stuck together but B is about to slip. F is pulling the mass of both. So the result of whatever forces are acting on B is a horizontal force giving to B exactly the acceleration that F gives to A+B. With this in mind, draw your diagram being very careful with sin and cos and where to put right angles, these were some of the mistakes made before.

If you do all that, you should be in a good position to solve it.

 

[haruspex]

Now for the math:
When stationary - Block A

ƩF_y = m_Aa_A = 0 = -49N + N_A

What about N_AB and F_fAB?

When F is applied - Block A

ƩF_x = m_A × -a_x = F cos30 + F_fa

Are you treating F_fa as negative? (You drew the arrow oppositely to F's)

ƩF_y = m_A × a_y = 0 = N_a - 49 + F sin30

Again, what about N_AB and F_fAB? As I suggested, it's simpler to treat A and B combined instead of A in isolation.

When F is applied - Block B

ƩF_x = -2a_x cos15 + (m_B × g × sin15) ± F_fAB

In your diagram you have x perpendicular to the slope and y parallel to it.
Is g acting in the direction of a or against it? Shouldn't you write
ƩF_x = 2a_x cos15 = -(m_B × g × sin15) + (friction term)
(Which way is friction acting here? )

So this will keep block B on block A so i guess the answer is F > 44.5N ?

I get a little over 60N.

 

[TSny]

If you consider the forces on B alone, you will be able to determine the numerical value of the acceleration of the system that will put B on the verge of slipping.

See if you can complete the free body diagram of block B that I have attached here. I leave it to you to identify the forces (blue vectors) and the appropriate angles.

As pointed out before, it will be best to choose the x-axis horizontally and the y-axis vertically.

Since we are assuming that B is on the verge of slipping there is a relationship between the friction force acting on B and the normal force acting on B.

What can you deduce from $\sum$F_y = ma_y ?

Then, what can you get from $\sum$F_x = ma_x ?

 

[sandman203]

ok one question.. when calculating force of friction for block A do i need to take into account the friction created from accelerating block B to the left?

 

[TSny]

You can find the friction force that the floor exerts on A by considering A and B together as one object, drawing a free body diagram for that object, and applying the 2nd law.

[EDIT: If you are referring to the normal force that A exerts on B, then you will be able to get that from applying $\sum$F_y = ma_y to the diagram for B alone( if you choose the y-axis vertical).]

 

[sandman203]

I give up guys, I've been working at this for the last 2 days (approx 16 hours) and i just can't for the life of me figure this one out? Should i drop out of my university course if i can't solve this? i feel like I am going einstein-style insane over this...

 

[sandman203]

Ok just say i have

ƩF_x = 2a_xcos15 = F_fb + 2×9.8sin15

Am i right in subbing in 0.6 x N_AB into F_fb? I guess I am having trouble on where to use 0.6 x N_AB in the equations

 

[sandman203]

Ok guys, last attempt at answering this Q before i /wrists

SO here are the 4 equations after applying F

Block AB:
ƩF_yAB = F sin 30 + N_AB - m_AB×g

∴ N_AB= m_AB×g - F sin30

ƩF_xAB = - Fcos30 +F_fA = - m_AB×a_x

Block B:

ƩF_xB = 2×-a_xcos15 = 2×g×sin15 - F_fb

ƩF_yb = 0 = N_B - 2×a_xsin15 - m_b×g

∴ N_B = 2×a_xsin15 + m_b×g×cos15 (Am i right in saying the acceleration of ax causes downward force on block B hence increasing the size of the normal?

→ F_fB = 0.6 (2a_xsin15 + 2×g×cos15)

from here i get a_x = -3.877478

so

F_xAB = Fcos30 + F_fAB = m_AB × -a_x → (1)

Now F_fA = 0.6 (7×g - F sin30) → (2)

after subbing 2 into 1 i get F = 58.577N not quite 60N but its close i guess! what u guys think?

 

[haruspex]

Block AB:
ƩF_yAB = F sin 30 + N_AB - m_AB×g

∴ N_AB= m_AB×g - F sin30

ƩF_xAB = - Fcos30 +F_fA = - m_AB×a_x

Block B:

ƩF_xB = 2×-a_xcos15 = 2×g×sin15 - F_fb

All looks good to here, assuming you are taking a_x as positive to the left.

ƩF_yb = 0 = N_B - 2×a_xsin15 - m_b×g

I feel you are confusing yourself by mixing forces and accelerations on the same side of the equation. Since the acceleration is to the left, its coefficient normal to the slope is downwards, so it is aided by g and opposed by the normal.

Am i right in saying the acceleration of ax causes downward force on block B hence increasing the size of the normal?

No, it's the other way around. The acceleration threatens to leave the block behind, so must be reducing the normal force.

from here i get a_x = -3.877478

If a_x is positive left, you should get a positive value.

 

[sandman203]

Im having trouble seeing how the acceleration to the left cause less normal force can someone please explain this to me, i really feel like I am going insane over this. please help!

 

[TSny]

Im having trouble seeing how the acceleration to the left cause less normal force can someone please explain this to me, i really feel like I am going insane over this. please help!

Please set up the equation $\sum$F_y = m_Ba_y for the diagram I posted for B. That will allow you to find the numerical value of the normal force that A exerts on B.

 

[sandman203]

ok so for block b ƩF_y = 2×-gcos15 = -Fsin15 + N_AB +2×a_xsin15??

 

[TSny]

Note $\sum$F_y = ma_y means

N_By + f_By + W_y = ma_y where W is the force of gravity, f_B is the friction force on B, and N_B is the normal force.

Keep in mind that in the free body diagram for B, we are choosing the y-axis vertically. Does block B have any acceleration in the vertical direction?

 

[sandman203]

ahh no it has none, otherwise it would lift off the block or go into block A. right?

 

[TSny]

Right. We are considering block B when it is just on the verge of slipping (but not yet slipping). So, yes, block B does not move vertically at all and a_y = 0.

Now, how would you express N_By in terms of N_B and the angle of 15 degrees?

How would you express f_By in terms of f_B and the angle of 15 degrees?

How would you express W_y in terms of m_B?

 

[sandman203]

ok so I've changed ƩFyb = 0 = NB + 2×axsin15 - mb×g

which decreases the normal.. i understand that now.. i didnt understand that before because i was thinking that only the x portion of the Force applied was acting on B. Now by realising the Y component of F also acts on B i can see how it reduces the normal.

So subbing into Fx of B i get an acceleration of 2.389s.

and subbing that into Fx equation for Block A i get 49.64N... is this correct?

 

[TSny]

ok so I've changed ƩFyb = 0 = NB + 2×axsin15 - mb×g

The zero on the left of the equation is correct and the -m_bg on the right is correct.

However, the y-component of N_B is not N_B. You're going to need a trig function to get the y-component. (Study the free body diagram.)

I don't know how you got the term 2×axsin15. This term should represent the y-component of the friction force f_B which will be f_B sin15.

You can then use f_B = μ_sN_B to write the y-component of f_B as μ_sN_Bsin15. So, your equation $\sum$F_y = 0 will be an equation with just one unknown, N_B. Therefore you can find the value of N_B.

Then, you can set up $\sum$F_x = ma_x for block B and find the acceleration.

So subbing into Fx of B i get an acceleration of 2.389s.

and subbing that into Fx equation for Block A i get 49.64N... is this correct?

These answers are not correct, but they are not too far off.

 

[sandman203]

why is it not Nb, the normal is perpendicular to the incline surface? 2×axsin15 is the vertical component of F acting on B? or does only the horizontal portion of F work on B? how does friction come into action at all on the y-axis? I've changed the co-ord system when dealing with B. should i have not done that or something?

 

[TSny]

why is it not Nb, the normal is perpendicular to the incline surface?

Remember, we are choosing the y-axis vertical; that is, the y-axis is perpendicular to the ground and not perpendicular to the incline. Look at the attached figure and you can see that N_By is not the same as N_B. Using trig on the right triangle you get

N_By = N_Bcos(15)

The figure also shows the y-component of the friction force, f_By. You can see that f_By = f_Bsin(15).

Since block B is on the verge of slipping, you can write f_B = μ_sN_B. So, that means

f_By = f_Bsin(15) = μ_sN_Bsin(15)

You also know that the y component of the weight is

W_y = -mg


There is no y-component of acceleration, so Newton's 2nd law gives $\sum$F_y = 0. That is,

N_By + f_By + W_y = 0.

Substitute for each of these terms using the red-colored expressions above and solve for N_B.

 

[sandman203]

Ok so using the above method i get Nb = 17.481, subbing into fxb i get ax = 2.529m/s^2 then subbing into Fxab i get 50.48N