# Dynamics Question

1. May 15, 2008

### looi76

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
F = m.a
Force = Change in momentum / Time

3. The attempt at a solution
Don't know how to start!...
(a) The car and the trailer are the pairs?

Last edited by a moderator: May 3, 2017
2. May 15, 2008

### Hootenanny

Staff Emeritus
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. May 15, 2008

### looi76

Hi Hootenanny

The pair is the gravity vs the car + the trailer?

4. May 15, 2008

### Hootenanny

Staff Emeritus
Sorry, I misread the question myself . The question gives you the action force,
What is the reaction force exerted on the car? What exerts this force?

5. May 15, 2008

### looi76

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. May 15, 2008

### Hootenanny

Staff Emeritus
Yes, so the other force is the reaction force of 150N exterted on the car by the trailer (in the opposite direction).

7. May 15, 2008

### looi76

(b)(i)
Force on the Horizontal Component:
$$150 - 20 = 130N$$

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

Resultant Force:
$$c^2 = a^2 + b^2$$
$$x^2 = 130^2 + 1863.9^2$$
$$x = \sqrt{130^2 + 1863.9^2}$$
$$x = 1861.4N$$

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

$$\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}}$$

8. May 15, 2008

### Hootenanny

Staff Emeritus
Good.
Hmm, are you sure that the trailer's weight is the only force acting in the vertical direction?

9. May 15, 2008

### looi76

Do you mean the 20N resistive force must also be used for calculating the vertical component force?

10. May 15, 2008

### Hootenanny

Staff Emeritus
No, the trailer is in contact with the ground so there will be a ...

11. May 15, 2008

### looi76

Frictional Force, but I don't think I will be able to calculate it with the details given!

12. May 15, 2008

### Hootenanny

Staff Emeritus
Normal force?

13. May 15, 2008

### looi76

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!

14. May 15, 2008

### Hootenanny

Staff Emeritus
If the car is on a smooth road, will the trailer accelerate in the vertical direction?

15. May 15, 2008

### looi76

$$m = 190kg \ , \F = 130N$$

$$a = \frac{F}{m} = \frac{130}{190} = 0.684ms^{-2}$$

Is this the acceleration of the trailer?

16. May 15, 2008

### Hootenanny

Staff Emeritus
Looks good to me

17. May 15, 2008

### looi76

By the acceleration of the trailer I should be able to find out part(b) (i) & (ii)?

18. May 15, 2008

### Hootenanny

Staff Emeritus
You have just answered (b)(i) and (ii).