Ramps and Newton's Second Law

  • Thread starter swede5670
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  • #1
swede5670
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Homework Statement


The 60 kg block starting at rest is pushed 7.8 m up a ramp at an angle θ = 26.5° to the horizontal in 15 s. If the coeficient of kinetic friction is 0.17:
What is the acceleration of block?
What is the force used to push the block up the ramp?

Homework Equations


Fnet = ma
Delta X = Vit + (1/2)at^2
Fg=ma
Fkinetic = μ * N

The Attempt at a Solution


I think I understand the first part,
7.8 = 0*15 + (1/2)(a)(15^2)
7.8 = 112.5a
.069=a

But the second part is what's giving me trouble
Do I need to make an Fnet = ma equation?
If I do I know that it would look like this
Force of Push - Force of friction = ma
Force of Push - Force of friction = (60)(.069)
Can I substitute in this equation? Fkinetic = μ * N for Fk?
Force of Push - (μ * N) = 60* .069
Fpush - (.17 * N) = 4.14

But what is my N force? I think that it would just be mass times acceleration of gravity and in that case:
Fpush - (.17 * (60 * 9.81)) = 4.14
Fpush - 100.06200 = 4.14
Fpush = 104.20200N

I'm not sure if this is right, I appreciate your help in advance.
 

Answers and Replies

  • #2
bowma166
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You did the first part right.

For the second part, you did everything right except your calculation of the normal force. Try drawing a force diagram of the block. On an inclined ramp, gravity doesn't act perpendicular to the plane of motion.
 
  • #3
swede5670
78
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Do I just need to find the horizontal component of gravity?
 
  • #4
bowma166
30
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Not the horizontal component, though you do need to break gravity into components. You need to find the component of gravity that is perpendicular to the ramp.
 
  • #5
swede5670
78
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So then do I just do
Cosine (26.5) = A/H
and in this case adjacent is the force I am looking for and H is gravity
Is the gravity Fg? so I can substitute Fg=ma
60 * 9.81 = Fg

Then I have
Cosine (26.5) * (60 * 9.81) = A

Is this correct?
 
  • #6
swede5670
78
0
Alright well that's wrong and I'm not sure why
Cos(26.5) * H = A
Cos(26.5) * (60 x 9.81) = A
which ends up being 118.71N
When I plug that in I get
Fpush - (.17 x 118.97) = 4.14
Fpush = 24.36
And I this doesn't work, so I'm not sure what's going wrong
 

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