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Applying Newtons laws

  • #1
I am having a very hard time understanding these two problems I have attached. I have read the chapter in our text over and over again, but do not see the correlation or how to exactly set the problems up. I recently had surgery on my writing hand and missed a large portion of this class, so I need to understand this. If anyone can thoroughly explain these problems, I would greatly appreciate it. Thank you.
 

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  • #2
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I am having a very hard time understanding these two problems I have attached. I have read the chapter in our text over and over again, but do not see the correlation or how to exactly set the problems up. I recently had surgery on my writing hand and missed a large portion of this class, so I need to understand this. If anyone can thoroughly explain these problems, I would greatly appreciate it. Thank you.
Welcome to physics forums!!!

You are supposed to follow the template provided for help with homework problems, and you are supposed to tell us your thoughts and analysis of the problem so far, so we can get you pointed in the right direction. So,...what do you think?

Chet
 
  • #3
Thank you and my apologies!

For the first problem, I understand that the ball is moving at a horizontal with uniformed circular motion, which is arad=V2/R. The Fnet=marad=mv2/R. The [itex]\sum[/itex]Fy=0 I think and [itex]\sum[/itex]Fx=marad? I'm not exactly sure about that or how to use the height depicted as y.

For the second problem, the box starts at rest so the initial velocity is 0 and i'm looking for the final velocity. It's also being pushed with a constant force F0. [itex]\sum[/itex]Fy= n + (-mg)= 0, so n=mg, but Im not sure how to write the sum of the forces in the x direction considering friction or where to go after that.
 
  • #4
19,796
4,044
Thank you and my apologies!

For the first problem, I understand that the ball is moving at a horizontal with uniformed circular motion, which is arad=V2/R. The Fnet=marad=mv2/R. The [itex]\sum[/itex]Fy=0 I think and [itex]\sum[/itex]Fx=marad? I'm not exactly sure about that or how to use the height depicted as y.
This is a correct assessment. Now the problem boils down to a geometry problem relating r to R and y. Draw a cross sectional diagram of the hemisphere.
For the second problem, the box starts at rest so the initial velocity is 0 and i'm looking for the final velocity. It's also being pushed with a constant force F0. [itex]\sum[/itex]Fy= n + (-mg)= 0, so n=mg, but Im not sure how to write the sum of the forces in the x direction considering friction or where to go after that.
This is a good start also. Do you know the relationship between the tangential frictional force in the x direction, and normal force, and the coefficient of kinetic friction?

Chet
 
  • #5
I don' know how to do that for number one. I also don't understand what your asking for number two. This is extremely frustrating. I don't have a clue where to look for this either.
 
  • #6
19,796
4,044
I don' know how to do that for number one. I also don't understand what your asking for number two. This is extremely frustrating. I don't have a clue where to look for this either.
I'm sorry my answer frustrated you. That was not my intention. By drawing a cross section diagram of the sphere, I came up with the following geometric relationship (involving a right triangle):
[tex](R-y)^2+r^2=R^2[/tex]
To do the next part of the problem, it is essential that you draw the same cross section diagram. Focus on the small ball as a free body. What is the force exerted by the hemisphere on the ball, and in what direction? What is the force that gravity is exerting on the ball, and in what direction? Are there any other forces acting on the ball? Write the force balance equations for the ball in the horizontal and vertical directions.

Now, for problem 2:

Draw a diagram of the system, and focus on the block as a free body. Identify the forces that are acting on the block. What are they (i.e., list them for us)? (hint: there are three forces).

Please get back with us with your answers, and we can continue.
 
  • #7
I don't know what a cross section diagram is. I don't know how you got that equation for number one.

For number 2, there are four forces not three right? Normal, Weight, Force, and Friction.
 
  • #8
19,796
4,044
I don't know what a cross section diagram is. I don't know how you got that equation for number one.
It's like slicing the hemisphere with a vertical plane, and looking at the resulting semicircular cut. Have you ever sliced an orange in half? You slice the orange in half, and look directly in at each half (perpendicular to the cut).
For number 2, there are four forces not three right? Normal, Weight, Force, and Friction.
There are three forces: weight, applied force, force from contact with plane. The contact force with the plane can be resolved into two components: Normal (vertical) and Frictional (horizontal). How is the frictional component related to the normal component N? Now do force balances on the block in the horizontal and vertical directions, and let us see what you get.

Chet
 
  • #9
Okay, but I still have no idea where you get the equation from that. Could you upload a picture of what you did?

Friction divided by normal is equal to the coefficient of friction right? Vertically you have normal - weight = 0 and horizontally you have the pushing force - the friction force but I don't know what that would be equal too.
 
  • #10
The forums are not loading correctly for me, so it will not let me input equations, but the sum of the forces in the x direction are = to F0 - (coefficient of friction * normal force) = mass * acceleration. Then I have to solve that for acceleration.
 
  • #11
19,796
4,044
The forums are not loading correctly for me, so it will not let me input equations, but the sum of the forces in the x direction are = to F0 - (coefficient of friction * normal force) = mass * acceleration. Then I have to solve that for acceleration.
The normal force is the weight of the block, mg, so the frictional force is mgμ0(1-x/L). So, from your word equation:

[tex]m\frac{dv}{dt}=F_0-mgμ_0(1-x/L)[/tex]

A trick to solving an equation like this is to multiply both sides of the equation by v, which is the same as dx/dt:

[tex]mv\frac{dv}{dt}=\frac{m}{2}\frac{dv^2}{dt}=F_0\frac{dx}{dt}-mgμ_0(1-x/L)\frac{dx}{dt}[/tex]

If you cancel out the dt's, you then get

[tex]\frac{m}{2}d(v^2)=F_0dx-mgμ_0(1-x/L)dx[/tex]

Then you integrate from x = 0 to x = L. The integral of the first term on the right hand side is the work done by the external force. The integral of the second term on the right hand side is the work done against friction. The integral on the left hand side is the change in kinetic energy of the mass. So this problem could also have been done (probably more simply) directly using "conservation of energy," rather than by a force balance.

Chet
 
  • #12
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[tex]\frac{m}{2}d(v^2)=F_0dx-mgμ_0(1-x/L)dx[/tex]

The integral on the left hand side is the change in kinetic energy of the mass.

Chet
Hi Chet...

I would like to understand a little bit of maths .

Is ∫d(v2) = v2 just like ∫dv = v ?

Is v2 treated like an ordinary variable ? Doesn't the power in 'v' make any difference ?

Thanks
 
  • #13
arildno
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I'll pop in:
1. "Is v2 treated like an ordinary variable ?"
In this case, yes.
2. " Doesn't the power in 'v' make any difference ?"
Not when we choose to let v^2 be our variable, rather than, say, "v".

Think in terms of the chain rule.

Suppose you have the function f(v)=F(u(v)), u=v^2

The quantity dF/du is still a meaningful quantity, isn't it?
:smile:
 
  • #14
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Hi Chet...

I would like to understand a little bit of maths .

Is ∫d(v2) = v2 just like ∫dv = v ?
Sure. Just think of ∫dU = U, where U = v2

Is v2 treated like an ordinary variable ? Doesn't the power in 'v' make any difference ?

Thanks
Yes. It is treaded like an ordinary variable. The power of v doesn't make any difference.
 
  • #15
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Thanks arildno and Chet
 

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