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Dynamics Question

  • Thread starter looi76
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Homework Statement


A car pulls a trailer with a force of 150N. This is shown in the diagram.
http://img91.imageshack.us/img91/4554/402ia2.png [Broken]
(a) According to Newton's third law, forces exist in pairs.
(i) What is the other force that makes up the pair?
(ii) Write down the size and direction of this force?

(b) The mass of the trailer is 190kg. As it sets off from rest, there is a resistive force of 20N.
(i) Calculate the size and direction of the resultant force on the trailer.
(ii) Calculate the initial acceleration of the trailer.

(c) The force from the car on the trailer is maintained at 150N.
When the car and trailer are traveling at constant speed.
(i) What is the size of the resultant force on the trailer.
(ii) Write down the size and direction of the resultant force on the trailer.

Homework Equations


F = m.a
Force = Change in momentum / Time

The Attempt at a Solution


Don't know how to start!...
(a) The car and the trailer are the pairs?
 
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Answers and Replies

  • #2
Hootenanny
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Hi again looi76,

Question (a)(i) is asking which force is the action force (as opposed to the reactionary force). In other words, which force was applied to then create the reaction force.
 
  • #3
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Hi again looi76,

Question (a)(i) is asking which force is the action force (as opposed to the reactionary force). In other words, which force was applied to then create the reaction force.
Hi Hootenanny :smile:

The pair is the gravity vs the car + the trailer?
 
  • #4
Hootenanny
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Hi Hootenanny :smile:

The pair is the gravity vs the car + the trailer?
Sorry, I misread the question myself :redface:. The question gives you the action force,
A car pulls a trailer with a force of 150N
What is the reaction force exerted on the car? What exerts this force?
 
  • #5
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The reaction force exerted by the car is the pull of the trailer and the force exerted by the trailer on the car is 150N ?
 
  • #6
Hootenanny
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The reaction force exerted by the car is the pull of the trailer and the force exerted by the trailer on the car is 150N ?
Yes, so the other force is the reaction force of 150N exterted on the car by the trailer (in the opposite direction).
 
  • #7
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(b)(i)
Force on the Horizontal Component:
[tex]150 - 20 = 130N[/tex]

Force on the Vertical Component:
[tex]m = 190kg \ , \ gravity = 9.81ms^{-2}[/tex]
[tex]F = m.a[/tex]
[tex]F = 190 \times 9.81[/tex]
[tex]F = 1869.3N[/tex]

Resultant Force:
[tex]c^2 = a^2 + b^2[/tex]
[tex]x^2 = 130^2 + 1863.9^2[/tex]
[tex]x = \sqrt{130^2 + 1863.9^2}[/tex]
[tex]x = 1861.4N[/tex]

Is this right? and how can I get the direction(angle)? Can I get it by using sine rule?

[tex]\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}}[/tex]
 
  • #8
Hootenanny
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(b)(i)
Force on the Horizontal Component:
[tex]150 - 20 = 130N[/tex]
Good.
Force on the Vertical Component:
[tex]m = 190kg \ , \ gravity = 9.81ms^{-2}[/tex]
[tex]F = m.a[/tex]
[tex]F = 190 \times 9.81[/tex]
[tex]F = 1869.3N[/tex]
Hmm, are you sure that the trailer's weight is the only force acting in the vertical direction?
 
  • #9
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Do you mean the 20N resistive force must also be used for calculating the vertical component force?
 
  • #10
Hootenanny
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Do you mean the 20N resistive force must also be used for calculating the vertical component force?
No, the trailer is in contact with the ground so there will be a ...
 
  • #11
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No, the trailer is in contact with the ground so there will be a ...
Frictional Force, but I don't think I will be able to calculate it with the details given!
 
  • #12
Hootenanny
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Frictional Force, but I don't think I will be able to calculate it with the details given!
Normal force?
 
  • #13
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(b)(i)
Force on the Vertical Component:
[tex]m = 190kg \ , \ gravity = 9.81ms^{-2}[/tex]
[tex]F = m.a[/tex]
[tex]F = 190 \times 9.81[/tex]
[tex]F = 1869.3N[/tex]
Isn't this the normal force? because they didn't mention the percentage of the frictional force so that I subtract it from it. Sorry, if I'm being stupid!:cry:
 
  • #14
Hootenanny
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Isn't this the normal force? because they didn't mention the percentage of the frictional force so that I subtract it from it. Sorry, if I'm being stupid!:cry:
If the car is on a smooth road, will the trailer accelerate in the vertical direction?
 
  • #15
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If the car is on a smooth road, will the trailer accelerate in the vertical direction?
[tex]m = 190kg \ , \F = 130N[/tex]

[tex]a = \frac{F}{m} = \frac{130}{190} = 0.684ms^{-2}[/tex]

Is this the acceleration of the trailer?
 
  • #16
Hootenanny
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[tex]m = 190kg \ , \F = 130N[/tex]

[tex]a = \frac{F}{m} = \frac{130}{190} = 0.684ms^{-2}[/tex]

Is this the acceleration of the trailer?
Looks good to me :approve:
 
  • #17
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By the acceleration of the trailer I should be able to find out part(b) (i) & (ii)?
 
  • #18
Hootenanny
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By the acceleration of the trailer I should be able to find out part(b) (i) & (ii)?
You have just answered (b)(i) and (ii).
 

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