Friction Force on Block on Table in Pulley System Homework

[allsmiles]

Pulley Systems! HELP!

Homework Statement

The three blocks in the diagram are released from rest and accelerate at the rate of 1.5m/s^2. What is the magnitude of the frictional force on the block that slides horizontally?

i don't know why my picture isn't attaching but basically the image is of a table and on it is 1 block(4kg), on the left dangling off the table with a pulley is another block (2kg), and the same on the right side except with a mass of 4kg

Homework Equations

Ff=MFn
f=ma

The Attempt at a Solution

so...i am very confused, i believe that I am trying to find the frictional force of the block on the table
i thought that the force of gravity and normal force cancel...and so what's left is the tension from the left and right, as well as the frictional force from the right...
the total force of the system is equal to the (total mass)(acceleration), which is 15N
now what? what do i do?!

 

[pgardn]

Homework Statement

The three blocks in the diagram are released from rest and accelerate at the rate of 1.5m/s^2. What is the magnitude of the frictional force on the block that slides horizontally?

i don't know why my picture isn't attaching but basically the image is of a table and on it is 1 block(4kg), on the left dangling off the table with a pulley is another block (2kg), and the same on the right side except with a mass of 4kg

Homework Equations

Ff=MFn
f=ma

The Attempt at a Solution

so...i am very confused, i believe that I am trying to find the frictional force of the block on the table
i thought that the force of gravity and normal force cancel...and so what's left is the tension from the left and right, as well as the frictional force from the right...
the total force of the system is equal to the (total mass)(acceleration), which is 15N
now what? what do i do?!

The frictional force opposes the motion of the box. Which way is the block on the table accelerating?

 

[allsmiles]

im guessing towards the right...with the block with the most mass?

 

[pgardn]

im guessing towards the right...with the block with the most mass?

So the frictional force is to the left? correct? What else besides the frictional force is acting on the block(on the table) to the LEFT?

 

[allsmiles]

the tension from the 2kg block dangling off the pulley to the left?

 

[pgardn]

the tension from the 2kg block dangling off the pulley to the left?

Yes indeed. So those two forces are acting to the LEFT. One Force is known, the other is not, friction. What forces are acting on the block to the RIGHT?

 

[allsmiles]

One Force is known, the other is not, friction.

how is tension known?

and from the left is the tension from the left block...the force of gravity and normal force cancel out...

 

[allsmiles]

btw thanks for taking the time to help me out, much appreciated!:)

 

[pgardn]

how is tension known?

and from the left is the tension from the left block...the force of gravity and normal force cancel out...

The block on the left must accelerate up. So T beats out mg by ma for the left, yes? In other words, Tension acting up on the 2kg mass is larger than mg acting down on the 2kg mass because the 2kg must accelerate up if the 4kg block accelerates to the right.

So T - 2g = 2a... you know g and a...

You really are best off doing this using free body diagrams and keep a strict accounting of all forces and the net force. Have you used these before?

 

[allsmiles]

oh wow
i never knew there was a tension equation...
i do have a free body diagram...and so if i have the tension(and they are equal on either side of the block, right?) and my only left unknown is friction...do i just say Ff+T=ma?
and then if they are equal on either side...dont they cancel??

 

[pgardn]

oh wow
i never knew there was a tension equation...
i do have a free body diagram...and so if i have the tension(and they are equal on either side of the block, right?) and my only left unknown is friction...do i just say Ff+T=ma?
and then if they are equal on either side...dont they cancel??

There is not a tension equation. Tension is considered a force. Look at this FBD... and see if it is understandable how I got T - mg = ma or T = mg + ma

This is a FBD for the 2kg mass. There is a net force on it. It must be accelerating up. The net force is just ma

 

[allsmiles]

ohhh ok i get itt
so then the net force of that block is equal to ma, and the net force is equal to only the force of friction? because the tensions cancel each other out?
but that can't be right...the answer is too easy that way...

 

[pgardn]

ohhh ok i get itt
so then the net force of that block is equal to ma, and the net force is equal to only the force of friction? because the tensions cancel each other out?
but that can't be right...the answer is too easy that way...

no...

Can you draw a FBD for the 4 kg block on the right? and then give me the equation you used to find this T on the right? its diff from the T on the left otherwise the 4 kg block on the table would not accelerate to the right. The T on the right must be bigger than the T on the left.

 

[allsmiles]

OMG I GOT IT!
ok:
Ft(from 1)=ma+fg(of 1)
Ft(from 3)=fg(of 3)-ma

Ff(of 2)=Ft3-Ft1-ma
plllllzzz tell me I am right! lol i hope you understood what i wrote

 

[pgardn]

Does it make sense that the tension on the block on the left is T = 2(10) + 2(1.5) ?

 

[allsmiles]

yup! so it's right then?
thank you SO much...could you just answer one more question for me, really quickly, i just need to clear up something

 

[pgardn]

OMG I GOT IT!
ok:
Ft(from 1)=ma+fg(of 1)
Ft(from 3)=fg(of 3)-ma

Ff(of 2)=Ft3-Ft1-ma
plllllzzz tell me I am right! lol i hope you understood what i wrote

What are 1 and 3? Is 1 the block on the left, the 2kg block and 3 is the block on the right, the 4 kg block? and I guess 2 is the block on the table?

 

[allsmiles]

oh yeh sry but yeh
1:the 2kg dangling off the left of table,
2: the 4kg, main one on the table
3: the 4kg, dangling off the right of the table

 

[pgardn]

oh yeh sry but yeh
1:the 2kg dangling off the left of table,
2: the 4kg, main one on the table
3: the 4kg, dangling off the right of the table

So it looks like you got it. But please, please look at my FBD for the block on the table. You need to be able to do this accurately because some of these problems can be cumbersome if you dont. I have examined all the forces in the x and y directions and then from my FBD's I have written some equations. Tell me you get this first.

Fn is the normal force.

 

[allsmiles]

ok...you lost me but i think i got it so wutever lol
thanks again! but please, just one more question
could you look at this one?
https://www.physicsforums.com/showthread.php?t=414563

 

[pgardn]

ok...you lost me but i think i got it so wutever lol
thanks again! but please, just one more question
could you look at this one?
https://www.physicsforums.com/showthread.php?t=414563

This one might be considered harder if the I lost you. I am going to draw the FB diagram of the forces on a block at rest on a plane. The plane must have friction or the block would accelerate. If you can't draw these you can be in real trouble later on... Then I got to go to bed. So you can look at this and try it on your own. I will probably be back tomorrow.

I have divided mg into sin angle m g and cos angle m g in the second picture so I could set them in line with the normal force and the force of friction.

A stands for angle, most of the times its written as theta. You must recognize this hopefully or no way we can get through the other one. You need to read about it if you don't know this. Have a good evening.

 

[allsmiles]

i know this and i know what theta is and i know how to draw an FBD diagram!
lol i did take calc and I am enrolled in grade 12 physics:)
thanks a lot once again!
gud night!<3