# Newtonian Mechanics Easy Question

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#1

I will have to develop two equations, one for m2, in order to find the acceleration vector, and one for m1 to find the force.

So the first equation:

[positive direction: to the right -------->]
sigma F(x) = m a(x)

20 - 0.3 ( 5 g ) = 5 a

therefore, a = 1.057 m s ^-2

The second equation for the m1 object:

Lets call the force F(2):

F(2) - 0.5 (2g) = 2 (1.057)

F(2) = 11.9 ms^-2 (3 sf)

This answer is wrong, I don't where I went wrong really. Any help would be appreciated. Thought this would refresh my basic mechanics skills.

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5 years ago
#2
(Original post by Daniel Atieh)
...
Draw two separate diagrams: one for the bottom block and one for the top block.

For the bottom block, the forces acting should be: weight, normal reaction, friction between the block and the surface, friction between the blocks and the applied force.

I assume the question means what is the horizontal force acting on
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#3
Draw two separate diagrams: one for the bottom block and one for the top block.

For the bottom block, the forces acting should be: weight, normal reaction, friction between the block and the surface, friction between the blocks and the applied force.

I assume the question means what is the horizontal force acting on
Thank you for taking part. How I can find the horizontal force acting on m2? Shouldn't it be on m1?
Like I find the acceleration using m2, and then I can find the horizontal force acting on m1. Btw, for m2, I will have to use the total mass of 5 Kg right?
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5 years ago
#4
(Original post by Daniel Atieh)
Thank you for taking part. How I can find the horizontal force acting on m2? Shouldn't it be on m1?
Like I find the acceleration using m2, and then I can find the horizontal force acting on m1. Btw, for m2, I will have to use the total mass of 5 Kg right?
You don't need to consider in this explicit manner.

Look at the bottom block. Horizontally, there is an applied force. There is also a frictional force due to the contact between the surface. There is another frictional force between the two blocks.

Sum these forces and you'll find the resultant horizontal force, acting on that is.
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#5
You don't need to consider in this explicit manner.

Look at the bottom block. Horizontally, there is an applied force. There is also a frictional force due to the contact between the surface. There is another frictional force between the two blocks.

Sum these forces and you'll find the resultant horizontal force, acting on that is.
aha. I got you. Now drawing the bottom block:

Even after I sum the forces, it's coming wrong. I believe I got something wrong in the frictional forces and Normal reaction :/
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5 years ago
#6
(Original post by Daniel Atieh)
...
Does the question say find the forces acting on or ? I'm actually asking, can't read the question.
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#7
Does the question say find the forces acting on or ? I'm actually asking, can't read the question.
I can't actually see either :/ But I think it says on m1. I didn't shoot this picture from my book so.
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5 years ago
#8
(Original post by Daniel Atieh)
I can't actually see either :/ But I think it says on m1. I didn't shoot this picture from my book so.
The frictional force acting between the surfaces isn't is it?

The reaction force is only here.
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#9
The frictional force acting between the surfaces isn't is it?

The reaction force is only here.
But there was another question with a similar situation, and it asked the normal reaction on the bottom object, and it turned out it should be (sum of masses x 9.81).
m2 has a mass of 3 Kg, and m1's mass = 5 Kg

The coefficient of kinetic friction between the surfaces is 0.3, and the coefficient of static friction between m1 and m2 is 0.5

I still don't know how to arrive to the answer
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5 years ago
#10
(Original post by Daniel Atieh)
...
The normal reaction on the bottom block on its lower surface is 5g. The normal reaction on the bottom block on its upper surface is 2g.

Use that and the coefficients of friction to find the forces acting.

That's as detailed help as I should be giving you really. If you still can't arrive at the answer, I'd recommend that you speak to your teacher.
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#11
The normal reaction on the bottom block on its lower surface is 5g. The normal reaction on the bottom block on its upper surface is 2g.

Use that and the coefficients of friction to find the forces acting.

That's as detailed help as I should be giving you really. If you still can't arrive at the answer, I'd recommend that you speak to your teacher.
I believe that I can reach to the answer with a few more steps, and not need to ask my teacher.

So I can find the acceleration using the bottom object. Then what keeps m1 moving with m2 is the static friction right? I can find it using F = ma
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5 years ago
#12
(Original post by Daniel Atieh)
I believe that I can reach to the answer with a few more steps, and not need to ask my teacher.

So I can find the acceleration using the bottom object. Then what keeps m1 moving with m2 is the static friction right? I can find it using F = ma
No. If you grapsed about 60% of the content required for this topic, you would be able to take a good stab at the question.

From what I see, you're missing a few fundamentals.
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#13
No. If you grapsed about 60% of the content required for this topic, you would be able to take a good stab at the question.

From what I see, you're missing a few fundamentals.
hmm. It took a number of posts in the beginning as the question wasn't clear, and I myself thought that it want us to work the horizontal force on the bottom object. Then it turned out it's for the top one. I just wanted to know where I have gone wrong in this, that's it.

I told you that I can find the acceleration using the bottom object (F=ma, where the forces are the pull of 20 N, and the kinetic friction 0.3 x 5g N, and the mass to be used is 5). Now that I have got the acceleration, I can use F=ma for the top one, and it turns out that the sigma F to the right = 2.1 N which is provided by the static friction (to the right) between m1 and m2.

What I want here is to know if this approach is correct or wrong.

Edit: maximum static friction is 9.8 N
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