Two blocks - one on a table and one hanging - acceleration with friction

[Hemmelig]

[SOLVED] Two blocks - one on a table and one hanging - acceleration with friction

Homework Statement

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that's a link to an image of the problem

Block A (mass 2.25kg) rests on a table top, it is connected by a horizontal cord passing over a light , frictionless pulley to a hanging block B (mass 1.30kg). The coefficient of kinetic friction between block A and the table is 0.450

After the blocks are released from rest, find the speed of each block after moving 3 cm.

Homework Equations

The Attempt at a Solution

I've calculated the normal force for both blocks

Block A = M1g = 2.25*9.81 = 22.07N
Block B = M2g = 1.30*9.82 = 12.7N

I'm a bit unsure on how to calculate the friction force fk for this one, if it had been just a single block sitting on a horizontal then it would've been the friction coefficient * mg
My plan was to use the formula : V^2 = V0^2 + 2a (x-x0)
to find the velocity after 3cm

The blocks will have the same acceleration and the same velocity, but i don't know how to calculate either due to the frictional coefficient.

If the friction wasn't involved, i could've used a=(M2g)/(m1 + m2 )

could someone please point me in the right direction ?

 

[Hootenanny]

I'm a bit unsure on how to calculate the friction force fk for this one, if it had been just a single block sitting on a horizontal then it would've been the friction coefficient * mg

You're on the right lines, but there are two blocks so you need to take into account both their masses.

As for what to do next, I'd make use of Newton's Second Law.

 

[Hemmelig]

Cheers for the reply

That's what i thought, so i tried going for friction force k = (friction coefficient * m1) + (friction coefficient * m2) = 15.64N

Is that correct ?

In that case, i know that the force holding it back is 15.64N , i know the total pull is normal force A + Normal force B is = 34,82N

Am i going in the right direction ?

I don't really know what to do next with Newtons second law

F=ma

According to the solution in the back of the book, the velocity is supposed to be 0.218m/s

 

[Hootenanny]

Cheers for the reply

That's what i thought, so i tried going for friction force k = (friction coefficient * m1) + (friction coefficient * m2) = 15.64N

Is that correct ?

Hang on sorry, scratch my previous post. I thought the two blocks were on top of each other, which is not the case. Since only block A experiences friction, you only need to calculate the frictional force for this block. For this bit you can ignore the second block and treat block A in isolation. Do you follow?

 

[Hemmelig]

Ah, in that case, i'll end up with a frictional force fk that is:

fk=frictional coefficient * m1g = 9.93

I still have no idea what to do next though :/

The magnitude of F is the sum of both blocks, right ?

So basicly F = (frictional coefficient * M1g) + (m2g)

F = 22.68N

Correct ?

 

[Hootenanny]

Ah, in that case, i'll end up with a frictional force fk that is:

fk=frictional coefficient * m1g = 9.93

Correct

The magnitude of F is the sum of both blocks, right ?

So basicly F = (frictional coefficient * M1g) + (m2g)

F = 22.68N

Correct ?

Not quite, but you have the right idea. Now can you start by writing down the sum of the forces acting on block A? Don't try and put numbers in, just use words or symbols.

 

[Hemmelig]

The sum of forces working on Block A is:

the normal force n, which is the same as mg
fk = the frictional force that tries to hold it back
then you have the force from block B which is trying to pull block A

 

[Hootenanny]

The sum of forces working on Block A is:

the normal force n, which is the same as mg
fk = the frictional force that tries to hold it back
then you have the force from block B which is trying to pull block A

Correct, since Block A doesn't move in the vertical direction we can ignore the normal force. So now we have,

\sum F_A = T - \mu mg

Where T is the tension in the string. Do you agree? Can you do the same for the second block?

 

[Hemmelig]

I see

I'm on my way out, so i'll try when i get back in

Just one more thing, i tried doing this and i got the right acceleration, but i don't really know if I'm allowed to do it

I know the frictional force fk that is trying to hold the blocks back

I know the force that block B is trying to pull with

So i tried force block b = mg = 12.753N

Total force = force block b - frictional force = 12,753 - 9.93 = 2.823

F=ma
F=2.823
m=block a + block b = 2.25 + 1.30 = 3.55

a = F/m = 2.823 / 3.55 = 0.792

When i then use V^2 = V0^2 + 2a (x-x0)

I get V= 0.2179

Which is the correct answer

 

[Hootenanny]

I'm afraid your solution isn't correct. It's just shear luck that your answer came out close to the actual answer of 0.21833...m/s

 

[Hemmelig]

Correct, since Block A doesn't move in the vertical direction we can ignore the normal force. So now we have,

\sum F_A = T - \mu mg

Where T is the tension in the string. Do you agree? Can you do the same for the second block?

Hmm, I'm not sure what the second block will be

\sum F_B= N + (-mg) ?

 

[Hootenanny]

Careful, there is no normal reaction force acting on block B.

Which two forces are acting on block B? In what direction to the act in relation to the motion of B (i.e. in the same direction of motion or in the opposite direction)?

 

[Hemmelig]

Careful, there is no normal reaction force acting on block B.

Which two forces are acting on block B? In what direction to the act in relation to the motion of B (i.e. in the same direction of motion or in the opposite direction)?

Hmm, the two forces acting on B is it's mass and g. They are acting downwards, which is in the same direction of the motion of B

Right ?

 

[Hootenanny]

Hmm, the two forces acting on B is it's mass and g. They are acting downwards, which is in the same direction of the motion of B

Right ?

No, mass isn't a force and neither is g (g is acceleration due to gravity, which is not a force). It's weight is one force, but what is the other one? What is block B attached to? Look at my expression in post #8.

Drawing a free body diagram of block B may help.

 

[Hemmelig]

Hmm

Well, it's weight is trying to pull block B down, while Block A and the friction is basicly holding back.

 

[Hootenanny]

Well, it's weight is trying to pull block B down

Correct

while Block A and the friction is basicly holding back.

Correct, and this resistance to motion is transferred through the tension is the rope.

Can you now write an expression for the sum of the forces on the second block, as I did for the first?

 

[Hemmelig]

The tension in the rope connecting the blocks is \mu * m(a)g

So the sum of all forces working on block B is it's weight trying to pull it down, minus the tension of the rope ?

 

[Hootenanny]

The tension in the rope connecting the blocks is \mu * m(a)g

No it isn't, if it was, your previous method would be correct. The tension has to account for the frictional force, the weight of the block B and the acceleration of the two blocks.

So the sum of all forces working on block B is it's weight trying to pull it down, minus the tension of the rope ?

This however, is correct. Therefore,

\sum F_B= m_Bg - T

Do you follow? Can you now apply Newton's second law to both these net forces?

 

[Hemmelig]

Hmm, i don't really know what to do next

It would be more of a guess than actual knowledge :/

 

[Hootenanny]

Hmm, i don't really know what to do next

It would be more of a guess than actual knowledge :/

A guess is better than nothing ! Newton's second law states that,

\sum \vec{F}=m\vec{a}

So, can you make an educated guess?

 

[Hemmelig]

A guess is better than nothing ! Newton's second law states that,

\sum \vec{F}=m\vec{a}

So, can you make an educated guess?

Well

My guess basicly brings me down the same route as before

Can i say:

Sum of all forces F= sum of Force(a) + sum of force(b) = 2.823N

Then i can just use F=ma to find a ? by dividing F with the total mass of the two blocks

But that's basicly what i did earlier on, which was wrong

 

[Hootenanny]

Well

My guess basicly brings me down the same route as before

Can i say:

Sum of all forces F= sum of Force(a) + sum of force(b) = 2.823N

Then i can just use F=ma to find a ? by dividing F with the total mass of the two blocks

But that's basicly what i did earlier on, which was wrong

No, unfortunately you can't do that. You must consider each block separately, so for the first block one would write,

\sum F_A = T - \mu m_A\cdot g = m_A\cdot a

Do you follow? Can you do the same for the second block?

 

[Hemmelig]

Hmm

<br /> \sum F_B = mg - T = m_B\cdot a

?

 

[Hootenanny]

Hmm

<br /> \sum F_B = mg - T = m_B\cdot a

?

Correct

 

[Hemmelig]

Hmm, so what do i do next ?

I don't know T and i don't know a

 

[Hootenanny]

Hmm, so what do i do next ?

I don't know T and i don't know a

Hmmm, you have a system of simultaneous equations, two equations with two unknowns. Solve.

 

[Hemmelig]

Ah, ofcourse

From equation a i have that T=12,753 - 1,3a

From equation b i have that T=2,25a + 9,93

I solve for T and get a=2,823/3,55 = 0,7952

I then put it into equation a and get t=11,71

I can then use V^2 = V0^2 + 2a (x-x0) to find the speed after 0.03m (is that the best equation to use ? )

I get v=0,2184307076

Thank you so much for your help

 

[Hootenanny]

I can then use V^2 = V0^2 + 2a (x-x0) to find the speed after 0.03m (is that the best equation to use ? )

Yup, that's the one that I would use

Thank you so much for your help

A pleasure .

As an aside, this method is applicable to any pulley question.

1. Identify the forces acting on each individual body.
2. Plug the results of (1) into Newton's Second Law, again for each individual body.
3. You know that if the string is light and inextensible the tension and acceleration must be the same for each body.
4. Solve for a and T.

 

[Hemmelig]

Great :)

Again, thank you so much