Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have a question about something I'm not sure about. I will first explain the story.

Image

If you calculate the force of a truck going downhill and applying its brakes:

  • The mass is $50000\text{ kg}$
  • The deceleration is $7\ \mathrm{m / s^2}$
  • And the angle is 7 degrees.

$F_1$ is the componenent of the gravity that is parallel to the to surface of the ground ( the F// or FII).

$$F_1 = 50000 *9.81* \sin(7^\circ) = 59777\text{ N (force downwards)}$$

The resulting force is $50000 \times 7 = 350000\text{ N}$ (using the deceleration value) (force downwards).

$$F_\text{brake} = 59777 + 350000 = 409777\text{ N (force upwards)}$$

If your truck goes up instead of down this time and then applies its brakes, do you need to do this?

$$F_\text{resulting} = F_\text{brake} + F_1 = 409777+59777 = 469554\text{ N}$$

Is it right to count $F_1$ twice, the first time in the F included in the brake and the second time separately?

share|improve this question
    
I edited your question to make it look a little better, but it's still not clear what you're really asking about or where some of these calculations come from. For instance, when you calculate $F_1$, where are you getting $50000\text{ N}$ from? What does $F_1$ represent, physically - the force applied by the brakes, or by the engine, or the parallel component of gravity? I think it would help your question a lot if you can explain more clearly what you're doing. –  David Z Oct 13 '11 at 18:55
    
I'm so sorry... –  user16987 Oct 13 '11 at 19:09
    
Don't be sorry - I'm not interested in an apology, I'm interested in helping you make your question better. Anyway: $F_1$ is the component of gravity acting down the slope, right? Then, if I understand you correctly, what you try next is using the given deceleration of $7 \mathrm{m/s^2}$ to calculate the amount of force that the brakes are able to apply? –  David Z Oct 13 '11 at 19:17
    
Ok :). I meant the componenent of the gravity that is parallel to the to surface (F// or FII or mg sin θ) . –  user16987 Oct 13 '11 at 20:15
add comment

1 Answer 1

You have an object of unit mass moving downhill, gravity acts in the direction of motion with a force G. The brakes act with a force B, so that the total acceleration is

$$a_1 = B + G$$

When the object is moving uphill, the gravity acts against the direction of motion, so the acceleration is

$$a_2 = B - G$$

The difference between the two situations is

$$ a_1 - a_2 = 2 G $$

So that if you want the acceleration when the truck is moving downhill, you add twice the gravitational force. The problem as you stated it is obscured by using large numbers, you should choose units to make everything be a small integer.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.