# Need help with this question

1. Sep 3, 2011

### [KNIGHT]

I need help with a physics question

1. The problem statement, all variables and given/known data
Figure shows 3 boxes with mass 50kg, 20kg & 30kg on a frictionless surface.

1. What is the horizontal force required to push the system with an acceleration of 1ms-1?
2. Find the Force exerted by 50kg on 20kg
3. Find the force exerted by 20kg on 30kg

2. Relevant equations

3. The attempt at a solution
I was able to find the answer for the 1st question by using F=ma. And the answer i got was F=100N. But i dont know how to do question 2 & 3. So please help.

Last edited: Sep 3, 2011
2. Sep 3, 2011

### stallionx

f->50<->20<->30

fbd's

f ->50<- fa

fa->20<-fb

fb->30

3. Sep 3, 2011

### [KNIGHT]

i dont understand how you have done it. So can you please explain it further?

4. Sep 3, 2011

### PeterO

Look at the 30kg box.

it is accelerating at 1 ms-2

What force is necessary to make that happen?

What is touching the 30kg mass, so what is providing the force?

EDIT: the diagram shows boxes of 50, 20 and 30 - your opening sentence says they are 50, 25 &30 ??

5. Sep 3, 2011

### [KNIGHT]

for some reason i dont understand this :( :( Can someone try re explaining again in the simplest form and answer those questions as if you are doing the sum? (but in the simplest way)

thanks for pointing it. corrected it

6. Sep 3, 2011

### PhanthomJay

You've got to draw Free Body Diagrams....Isolate the block you are investigating, identify the forces acting on it...and apply Newton's Laws. As PeterO has noted, the acceleration of each block is the same as the acceleration of the entire system of blocks.

7. Sep 3, 2011

### whoareyou

Think about action/reaction forces and Newton's Third Law

8. Sep 3, 2011

### [KNIGHT]

can you do the questions step by step. then i might be able to understand

9. Sep 3, 2011

### PeterO

The step by step solution to part 3, which you do BEFORE part 2 is. as I stated before;

Look at the 30kg box.

it is accelerating at 1 ms-2

What force is necessary to make that happen?

What is touching the 30kg mass, so what is providing the force?

There .. I have asked 2 questions - the answers to which give the answer to question 3.

What are your answers to those two questions?

10. Sep 3, 2011

### [KNIGHT]

By using F=ma
= 30x1
= 30N

20kg mass is touching it

Then is 30N the the force exerted by 20kg on 30kg? I that the answer?

Last edited: Sep 3, 2011
11. Sep 3, 2011

### PeterO

That is certainly part 3.

Now you need to look at Part 2 ...

12. Sep 4, 2011

### [KNIGHT]

If i have to redo the same question (all the parts) when the 3 boxes are on a rough surface of coefficient of static friction 0.6, how do i do it? Is it by using the equation F=miu.R? If so what should i substitute for F & R?

13. Sep 4, 2011

### PeterO

R will be the reaction force to the weight of what ever you were considering.
When you did part 1, you [correctly] just used a total mass of 100kg.

When you did part 3 you just used a mass of 30.

When you did part 2 you will have either used a mass of 20 or 50 depending which way you evaluated it.

F is the size of the friction force opposing what ever force you are calculating - the set of boxes, just 2 boxes, just 1 box; depends what you are analysing.

14. Sep 4, 2011

### [KNIGHT]

This is how i did part 2 (without friction)
F =ma
=20x1
= 20N

Then i did 2 parts with friction.

part 1
The total mass of the system = 100Kg.
g = 10Ms-2
Therefore mg = 1000N.
Since mg = R
R = 1000N
F=miu.R
=0.6X1000
= 600N
Frictional force = 600N

F=ma
F-600 = 100X1
F= 700N

part 3
Force exerted by 20kg on 30kg
Mass = 30kg
g = 10ms-2
mg = R = 300N

F= miu.R
= 0.6 X 300
= 180N

F=ma
F-180 = 30X1
F = 210N

Force exerted by 20kg on 30kg = 210N

Have i done it correctly

15. Sep 4, 2011

### PeterO

Your part 2 is incorrect [both times not surprisingly]

Without friction:
The net force on the 20 kg block is indeed 20N, but there are two blocks touching it.

The 50kg block pushes to the right, the 30kg pushes to the left.
The 30 kg block pushes left with 30N [Newtons 3rd law on the answer to part 3]
To get a net 20N force to the right, the 50N block must push with 50N.

[the fact that 30 goes with 30 and 50 goes with 50 is a bit of a co-incidence here]

The other way to do part 2 is a bit like you did Part 1.

In Part 1 you said F was accelerating a total of 100kg at 1 m/s^2 so F = ma gave 100N

Well the 50kg block is pushing against the other 2 blocks , of total mass 50Kg, and thus with a force of 50N.

You can fix up the Part 2 with friction with one of those methods now.

16. Sep 4, 2011

### [KNIGHT]

so part 2 without friction

from left to right
F=ma
F-30 = 20x1
F= 50N

Then is it ok?

And what about part 1 & 3 with friction?

17. Sep 4, 2011

### PeterO

Your with friction for 1 & 3 were fine. ANd that above is one way of doing part 2 - no friction.

18. Sep 4, 2011

### [KNIGHT]

Then how should i do part 2 with friction? Is it like this

F=miu.R
= 0.6 x (200N+300N)
= 300N

F=ma
F-300 = (20N+30N)x1
F = 800N

Last edited: Sep 4, 2011