Motion and Forces

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Homework Statement


A 3.54 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 10.6 N at an angle theta = 15.0° above the horizontal, as shown in the Figure. The coefficient of kinetic friction between the block and the floor is 0.08. What is the speed of the block 3.9 s after it starts moving?


Homework Equations


Xcomp= 10.6 cos 15
Ycomp= 10.6 sin 15

I don't know where to go from here :(
 

Answers and Replies

  • #2
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How about Newton's second law?

[tex] \sum_{i=1}^n F_i = ma[/tex]
 
  • #3
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so do I have to find the resultant vector or no?
 
  • #4
learningphysics
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so do I have to find the resultant vector or no?
Did you draw a diagram? Do [tex]\Sigma\vec{F} = ma[/tex] in the vertical and horizontal directions...
 
  • #5
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I don't know? I am so lost!
 
  • #6
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Okay so I have come up with the net force for the x and y directions...

Fxnet = Fx + Wx + Nx = Fx + 0 + Nx = max
Fynet = Fy + Wy + Ny = Fy - Wy + 0 = may
 
  • #7
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Is that set up right?
 
  • #8
learningphysics
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Is that set up right?
For the x-direction you wrote:

[tex]F_x + N_x = ma_x[/tex]

What is N_x ? Does the normal force act in the x - direction? what about the frictional force?

For the y-direction you wrote:

[tex]F_y - W_y + 0 = ma_y[/tex]

Hmmm... why didn't you put a normal force here?
 
  • #9
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For the x-direction you wrote:

[tex]F_x + N_x = ma_x[/tex]

What is N_x ? Does the normal force act in the x - direction? what about the frictional force?

For the y-direction you wrote:

[tex]F_y - W_y + 0 = ma_y[/tex]

Hmmm... why didn't you put a normal force here?
Nx doesn't act in the x direction so I put it as 0...
so I have...
Fx/m = ax

For the normal force in the y direction I put 0...
(Fy-Wy)/m = ay
 
  • #10
learningphysics
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Hmmm... you're missing forces.

List all the forces acting in the x-direction.

List all the forces acting in the y-direction.
 
  • #11
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Hmmm... you're missing forces.

List all the forces acting in the x-direction.

List all the forces acting in the y-direction.
Forces in the x-direction:

Tension
Normal
Fricton

Forces in the y-direction:

Weight
Normal
Friction
Tension
 
  • #12
learningphysics
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There's no normal force in the x - direction. There's no frictional force in the y-direction.

So in the x-direction... you only have the x-component of tension, and the friction... write out the F=ma equation in the x-direction.

In the y-direction... you have the y-component of tension, gravity and the normal force... write out the F=ma equation in the y-direction.
 
  • #13
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There's no normal force in the x - direction. There's no frictional force in the y-direction.

So in the x-direction... you only have the x-component of tension, and the friction... write out the F=ma equation in the x-direction.

In the y-direction... you have the y-component of tension, gravity and the normal force... write out the F=ma equation in the y-direction.
Oh okay...that was a dumb mistake I made...

Fxnet = Tx + Fx = ma
Fynet = Ty - Wy + Ny = ma
 
  • #14
learningphysics
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Cool.. try to substitute the values of tension, friction, weight etc... into your equations...
 
  • #15
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Cool.. try to substitute the values of tension, friction, weight etc... into your equations...
I got the friction and weight but I m drawing a blank on how to calculate the tension?
 
  • #16
learningphysics
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I got the friction and weight but I m drawing a blank on how to calculate the tension?
Actually you have the components of tension in your first post. :wink:
 
  • #17
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Actually you have the components of tension in your first post. :wink:
okay so the fx = 10.6 cos 15 is actually the tension force not the friction force?
 
  • #18
learningphysics
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okay so the fx = 10.6 cos 15 is actually the tension force not the friction force?
yes, 10.6 is the force the cord exerts... ie tension... so 10.6cos15 is the horizontal component... 10.6sin15 is the vertical component.
 
  • #19
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yes, 10.6 is the force the cord exerts... ie tension... so 10.6cos15 is the horizontal component... 10.6sin15 is the vertical component.
then how do I calculate the friction force. I know I have to use the Kinetic friction force of 0.08 but isn't there something else I have to do to it?
 
  • #20
learningphysics
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then how do I calculate the friction force. I know I have to use the Kinetic friction force of 0.08 but isn't there something else I have to do to it?
Frictional force = normal force * uk

uk is the coefficient of kinetic friction = 0.08
 
  • #21
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Frictional force = normal force * uk

uk is the coefficient of kinetic friction = 0.08
Thats what I thought but isn't the normal force in the y direction 0? So the frictional force would then be 0...that can't be right. I thought the normal force in the y firection was 0 because it lies in the y-axis.
 
  • #22
learningphysics
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Thats what I thought but isn't the normal force in the y direction 0? So the frictional force would then be 0...that can't be right. I thought the normal force in the y firection was 0 because it lies in the y-axis.
No normal force isn't 0... the net force is 0 in the y-direction (ie ay = 0, may = 0)
 
  • #23
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No normal force isn't 0... the net force is 0 in the y-direction (ie ay = 0, may = 0)
so I put...
Fxnet= (Tx + Fx)/m = a

but for normal force is that equal wcos theta?
 
  • #24
learningphysics
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so I put...
Fxnet= (Tx + Fx)/m = a
You mean this:

[tex]\Sigma\vec{F_x} = ma_x[/tex]

so substituting into the left hand side

[tex]Tcos(\theta) - friction = ma_x[/tex]


but for normal force is that equal wcos theta?
Don't assume that... write the equation for [tex]\Sigma\vec{F_y} = m\vec{a_y}[/tex]
 
  • #25
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why is it - friction and not +?

Ty - Wy + Ny = may
 

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