Motion & Forces: Find Block Speed After 3.9s

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Homework Help Overview

The discussion revolves around a physics problem involving a block being pulled on a horizontal surface with a force at an angle, while considering the effects of friction. The participants are tasked with determining the speed of the block after a specific time interval.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the need to analyze forces in both the x and y directions. There are questions about the setup of equations, the role of tension, and how to account for friction and normal forces.

Discussion Status

The discussion is active, with participants attempting to clarify their understanding of the forces involved and how to set up the equations correctly. Some participants express confusion about the normal force and friction, while others provide guidance on how to approach the problem.

Contextual Notes

Participants are navigating through the complexities of force components, frictional forces, and the implications of the block's motion on the normal force. There is an acknowledgment of the need to correctly identify forces acting in different directions and their contributions to the overall motion.

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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 :(
 
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How about Newton's second law?

[tex]\sum_{i=1}^n F_i = ma[/tex]
 
so do I have to find the resultant vector or no?
 
BuBbLeS01 said:
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...
 
I don't know? I am so lost!
 
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
 
Is that set up right?
 
BuBbLeS01 said:
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?
 
learningphysics said:
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
Hmmm... you're missing forces.

List all the forces acting in the x-direction.

List all the forces acting in the y-direction.
 
  • #11
learningphysics said:
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
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
learningphysics said:
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
Cool.. try to substitute the values of tension, friction, weight etc... into your equations...
 
  • #15
learningphysics said:
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
BuBbLeS01 said:
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
learningphysics said:
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
BuBbLeS01 said:
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
learningphysics said:
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
BuBbLeS01 said:
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
learningphysics said:
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
BuBbLeS01 said:
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
learningphysics said:
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
BuBbLeS01 said:
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
why is it - friction and not +?

Ty - Wy + Ny = may
 
  • #26
BuBbLeS01 said:
why is it - friction and not +?

Friction acts opposite the direction of motion... ie it tries to stop motion.

Ty - Wy + Ny = may

ok so

[tex]Tsin(\theta) - mg + N = 0[/tex]

So your two equations are:

[tex]Tcos(\theta) - friction = ma_x[/tex]
[tex]Tsin(\theta) - mg + N = 0[/tex]

I also know that friction is uk*N, and I can plug in T and theta so:

[tex]10.6cos(15) - \mu_k*N = ma_x[/tex]
[tex]10.6sin(15) - mg + N = 0[/tex]

The unknowns in this equation are N and ax... you have two equations 2 unknowns... you should be able to solve a...
 
  • #27
quick question...why is Fynet = 0?
 
  • #28
for finding I have N I have
10.6 sin - 34.73N = -N
can I rewrite that as
10.6 sin + 34.73N = N
so I don't have that -N?
 
  • #29
BuBbLeS01 said:
quick question...why is Fynet = 0?

because the block is sliding horizontally... if the tension in the cord is large enough then Fynet won't be zero... but then there won't be a normal force or friction either... the block will lift off the ground... I'll put it this way... the vertical tension in the cord isn't large enough to lift the block... so the block stays on the ground.

As long as the block is on the ground vertical acceleration is 0, hence Fynet = 0.
 
  • #30
BuBbLeS01 said:
for finding I have N I have
10.6 sin - 34.73N = -N
can I rewrite that as
10.6 sin + 34.73N = N
so I don't have that -N?

If you multiply both sides by -1... then you get

-10.6sin(15) + 34.73 = N
 

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