Two connected masses on separate inclines

[alex12342011]

1.

Homework Statement
See attachment please

Mass block 1= 20kg
Mass block 2= 30kg

Incline block 1= 40 degrees
Incline block 2= 60 degrees

Coefficient of kinetic friction block 1= 0.20
Coefficient of kinetic friction block 2= 0.30

Acceleration due to gravity= 9.81m/s^2Determine the acceleration and the tension n the rope. Assume pulley is mass less/frictionless and the rope is mass less.

Homework Equations

F=m*a
Ffriction= coefficient*Fnormal

The Attempt at a Solution

Drew fbd for each block

Found weight of each block using F=ma
Block 1 Fg= 196N
Block 2 Fg= 294N

Split them into components
Block 1
Fgx= 196sin40=126N
Fgy= 196cos40=150N

Block 2
Fgx= 294sin60=255N
Fgy=294cos60=147N

Fnormal = Fgy

Block1 Ffriction= 0.2(150)= 30N
Block2 Ffricion= 0.3(147)=44N

I am not sure which way it will accelerate. At first I thought it would accelerate to the right, however the answer says otherwise. To find which way it accelerates, I thought I would solve for the frictional force and the x component of gravity for each block. Then
compare them to see which is greater (I pretend as if there was no rope/force of tension)
Is this the correct way of finding what way it accelerates?

The answer told me it would accelerate left, so I continued knowing this and set my x axis
to make the acceleration in the x-axis positive.

Block 1 (20kg)

Net force= Fgx-Ffriction-Ftension
Net force= 126-30-Ftension
Net force= m*a
20a=126-30-Ftension
20a=96-Ftension
Ftension=96-20a

Block 2 (30kg)

Net force= Ftension-Fgx-Ffriction
Net force= Ftension-255-44
Net force= m*a
30a=Ftension-255-44
30a=Ftension-211
Ftension=30a+211

So now I set them equal to each other to find the acceleration?
96-20a=30a+211

-20a-30a=211-96

-50a=115
a=-2.3m/s^2
But this answer is not right? I am very confused. Any help is appreciated thanks!

Answer:
Acceleration= 1.1m/s^2 left
Force of tension= 180N

 

[SammyS]

1.

Homework Statement
See attachment please

Mass block 1= 20kg
Mass block 2= 30kg

Incline block 1= 40 degrees
Incline block 2= 60 degrees

Coefficient of kinetic friction block 1= 0.20
Coefficient of kinetic friction block 2= 0.30

Acceleration due to gravity= 9.81m/s^2


Determine the acceleration and the tension n the rope. Assume pulley is mass less/frictionless and the rope is mass less.

Homework Equations

F=m*a
Ffriction= coefficient*Fnormal

The Attempt at a Solution

Drew fbd for each block

Found weight of each block using F=ma
Block 1 Fg= 196N
Block 2 Fg= 294N

Split them into components
Block 1
Fgx= 196sin40=126N
Fgy= 196cos40=150N

Block 2
Fgx= 294sin60=255N
Fgy=294cos60=147N

Fnormal = Fgy

Block1 Ffriction= 0.2(150)= 30N
Block2 Ffricion= 0.3(147)=44N

I am not sure which way it will accelerate. At first I thought it would accelerate to the right, however the answer says otherwise. To find which way it accelerates, I thought I would solve for the frictional force and the x component of gravity for each block. Then
compare them to see which is greater (I pretend as if there was no rope/force of tension)
Is this the correct way of finding what way it accelerates?  No. if one mass is much less than the other, its acceleration might be large, but the force of gravity on a block of much larger mass might overwhelm the force of gravity on the block with small mass.

The answer told me it would accelerate left, so I continued knowing this and set my x axis
to make the acceleration in the x-axis positive.

Block 1 (20kg)

Net force= Fgx-Ffriction-Ftension
Net force= 126-30-Ftension
Net force= m*a
20a=126-30-Ftension
20a=96-Ftension
Ftension=96-20a

Block 2 (30kg)

Net force= Ftension-Fgx-Ffriction
Net force= Ftension-255-44
Net force= m*a
30a=Ftension-255-44   -255-44 = -299
30a=Ftension-211
Ftension=30a+211

So now I set them equal to each other to find the acceleration?
96-20a=30a+211

-20a-30a=211-96

-50a=115    203/(-50)=-4.06, which I bet is also wrong.
a=-2.3m/s^2  This means that if you assume motion to the left, the masses will slow down.
But this answer is not right? I am very confused. Any help is appreciated thanks!

Answer:
Acceleration= 1.1m/s^2 left
Force of tension= 180N

There should be two answers for acceleration. One if you assume motion is to the right, the other if you assume motion is to the left. This is because the frictional force is in a direction opposite the direction of the motion.

Since the given answer is that acceleration is to the left, both frictional forces should be toward the right, and I see that's what you did.

Assume that motion is to the right, so positive a is to the right & both frictional forces are to the left. If this results in negative acceleration, all that means is that if you assume motion to the right, the system will slow down. (I expect this direction will also give negative acceleration.)

 

[alex12342011]

There should be two answers for acceleration. One if you assume motion is to the right, the other if you assume motion is to the left. This is because the frictional force is in a direction opposite the direction of the motion.

Since the given answer is that acceleration is to the left, both frictional forces should be toward the right.

I don't quite understand why there are two answers for acceleration. Isn't there only one possible motion? If you assume it moves right, then shouldn't the net force acting on each block tell you otherwise?

Edit: Do you mean that you must assume the left/right motion of the whole system?

 

[Doc Al]

I am not sure which way it will accelerate. At first I thought it would accelerate to the right, however the answer says otherwise.

I think you were right the first time.

To find which way it accelerates, I thought I would solve for the frictional force and the x component of gravity for each block. Then
compare them to see which is greater (I pretend as if there was no rope/force of tension)
Is this the correct way of finding what way it accelerates?

To find which way it accelerates just guess and see if you're right. For example: You guessed left, but your answer came out negative. So that was wrong. Now assume it goes right and see if you can solve it.

 

[alex12342011]

Oh okay I understand now. Since my answer is negative, then I will just assume motion to the left. Then it should be positive and correct. Thank you!

 

[Doc Al]

Oh okay I understand now. Since my answer is negative, then I will just assume motion to the left.

You already assumed it went left and got a negative answer, so it goes right. You'll have to redo the calculation assuming it goes to the right.

 

[SammyS]

I edited my first post.

In short, I think if you assume motion is to the left, you will again get a negative acceleration.

(I know, I should probably do the calculation before making such a prediction.)

 

[SammyS]

Yup, I should have done the calculation before making such a quick prediction.

Assuming that motion is to the right, acceleration will also be to the right, but fairly small.

 

[alex12342011]

Oops I meant to say right sorry.

So assuming right...

Block 1 (20kg)

Fnet= (-30)+(-126)+Ft
20a = -156+Ft
Ft= 56+20a

Block 2 (30kg)

Fnet= (255)+(-44)+Ft
30a = 211+Ft
Ft= -211+30a


Set them equal

56+20a=-211+30a
20a-30a=-211-56
-10a=-267
a=26.7m/s^2


That seems pretty fast.

 

[alex12342011]

The answer is supposed to be 1.1m/s left? So that answer isn't possible?

 

[Doc Al]

Block 2 (30kg)

Fnet= (255)+(-44)+Ft

Ft will be in the negative direction.

 

[Doc Al]

The answer is supposed to be 1.1m/s left? So that answer isn't possible?

The direction is wrong.

 

[alex12342011]

Oops. Correcting Ft into the right direction, this is what I get:

211-30a=56+20a
-50a=-155
a=3.1m/s^2

Therefore a= 3.1m/s^2

The book however says 1.1m/s^2

, I don't think the books answer makes sense.

Thanks for the help.​

 

[Doc Al]

Oops. Correcting Ft into the right direction, this is what I get:

211-30a=56+20a
-50a=-155
a=3.1m/s^2

Therefore a= 3.1m/s^2

​

Recheck your work in arriving at your first equation here.​

 

[alex12342011]

211-30a=20a+156
-50a=-55
a=1.1

So finally... the acceleration is 1.1m/s^2

Subbing it in either previous equation to find tension

Block 1

Ft=20(1.1)+156
Ft=178

Double check with block 2

Ft=211-30(1.1)
Ft=178

Force of tension is 178N. 180N rounded up as book states.

Thanks for the help guys! ​

 

[SammyS]

The answer is supposed to be 1.1m/s left? So that answer isn't possible?

Yes it is possible. That's what I got.

See comments in red below.

Oops I meant to say right sorry.

So assuming right...

Block 1 (20kg)

Fnet= (-30)+(-126)+Ft
20a = -156+Ft
Ft= 56+20a   Ft= 156+20a

Block 2 (30kg)

Fnet= (255)+(-44)+Ft  This should be F_Net = (255) + (-44) + (-F_Tension)
30a = 211+Ft
Ft= -211+30a


Set them equal

56+20a=-211+30a
20a-30a=-211-56
-10a=-267
a=26.7m/s^2


That seems pretty fast.

 

[Doc Al]

Yes it is possible. That's what I got.

Show how you determined that it accelerates to the left.

 

[SammyS]

Show how you determined that it accelerates to the left.

That's for motion to the left and it does indeed accelerate to the left.

My prediction that it would decelerate in this case proved to be wrong.

Use the correction for the sign of F_Tension for block 2, in the quoted section of my previous post, (and use 156, not 56 in the F_Tension equation for block 1).

It does work!

 

[alex12342011]

That's for motion to the left and it does indeed accelerate to the left.

My prediction that it would decelerate in this case proved to be wrong.

Use the correction for the sign of F_Tension for block 2, in the quoted section of my previous post, (and use 156, not 56 in the F_Tension equation for block 1).

It does work!

But if you assume left and set the axes accordingly, then the friction would be negative and the component of gravity positive.

Block 1

assuming it goes left
Fnet=(126N)+(-30N)+Ft
20a=96N+Ft
Ft=-96+20a

Its not 156 anymore?

 

[SammyS]

But if you assume left and set the axes accordingly, then the friction would be negative and the component of gravity positive.

Block 1

assuming it goes left
Fnet=(126N)+(-30N)+Ft
20a=96N+Ft
Ft=-96+20a

Its not 156 anymore?

If you want a = 1.1 m/s², then assume they both go right.

Your Original Post had them both going left.

 

[SammyS]

Doc Al & alex12342011,

Sorry for the confusion I've caused! (I need to stop answering these when looking into a mirror! LOL)

When I said I got the answer "a = 1.1 m/s²", I didn't notice the "

" with it in alex's post (#15).

Yes, I did get a = 1.1 m/s², however, that acceleration was to the right assuming that the motion was also to the right. So, when I did this, I took the frictional forces to be toward the left.

The red comments I made in alex's post #9, in which he had motion to the right, gave me the answer: a = +1.1 m/s², where the positive direction is to the right.

Again, Sorry for the mix-up.


.​