# Inclined Plane Forces

1. May 9, 2006

### Ideologue

Please could you help me find some equations that can be used to calculate the power needed to accelerate up an inclined plane and also the power needed to travel up an inclined plane at a given speed?

Ignoring rolling resistance, aerodynamic drag, mechanical loss and the force needed to spin wheels, I need to know the force required for a bike to travel up the inclined plane.

Thus far I only have the formula below:

Force = weight X the sin of the angle of the inclined plane

That formula gives the force needed to move an object up the inclined plane, but I do not know how to incorporate the speed of the object or the acceleration of the object into it.

2. May 9, 2006

### lightgrav

Which Force is that? [gravity!]

Which Energy is associated with gravity?

How is Power related to Energy?

3. May 10, 2006

### Ideologue

It is ultimately the power required to ride a bike up an inclined plane (ignoring friction) that I seek to investigate.

I think the following equations are correct. Please comment on them either way:

Force required to accelerate up an inclined plane:

Force = weight X the sine of the angle X acceleration in meters squared

Force required to travel up an inclined plane at a given speed:

Force = weight X the sine of the angle X velocity

Are these accurate?

4. May 10, 2006

### Hootenanny

Staff Emeritus
This one is close.

Not so close. Go back to basics. What is Newton's first and second laws? A diagram of the forces acting would be useful in this case.

~H

5. May 11, 2006

### Ideologue

With ‘this is close’ I guess you mean completely wrong!

For acceleration up the inclined plane is this right:

F = mg sin(angle)

And for a constant speed up the slope:

F = mg sin(angle) + ma

?

6. May 12, 2006

### arunbg

P = Work/Time = Force*Distance/Time = Force*Velocity
Force required = mgsin(angle)
I think you can go from here.

Arun

7. May 12, 2006

### Hootenanny

Staff Emeritus
These two are the wrong way round! For constant speed, the driving force (up the plane) must equal the component of gravity pulling down the plane. Therefore For constant speed: $F = mg\sin\theta$.

To accelerate up the plane, you must first have a force equal to the component of gravity (or else you would accelerate down the plane!), then you must add the additional force required to accelerate up the plane at a m.s-2. Therefore To accelerate up the plane at a m.s-2: $F = mg\sin\theta + ma$.

Do you follow?

~H

8. May 12, 2006

### Ideologue

Thank you both for taking the time to post,

OK, I think I understand this now. So if I want to know how much force is required to accelerate a bike up a hill (ignoring friction and drag) from 0 to 10 miles per hour within two seconds then I should use the following equation:

f = mgsinθ + ma

If the inclined plane is at an angle of 20 degrees, the bike weighs 50kg and g = 9.81m/s squared and 10 miles per hour is the same as 4.4704 metres per second (and 0 to 4.4704 metres per second means that acceleration (a) = 2.2352 metres per second squared) then the following is correct:

F = 50 X 9.81 X 0.342020143 (being the sine of the angle) + (50 X 2.2352)

Answer = 279.520881 newtons.

?

9. May 12, 2006

### Hootenanny

Staff Emeritus
Yes, your correct. As long as your bike doesn't have a rider

~H

10. May 12, 2006

### Ideologue

lol, I did forget about the rider!

Many thanks for your help, I appreciate it.

11. May 12, 2006

### Hootenanny

Staff Emeritus
No problem! Mind you, 50kg is a heavy bike, I wouldn't like to ride it! Mines nice and light at about 10 kg and that's quite heavy for a bike!

~H

12. Jan 19, 2010

### G10slayer

I have been working on this same equation except I derived the equation F = weight*sine squared.

13. Jan 19, 2010

### G10slayer

(THIS IS FOR ABOVE)

friction is absent and speed is constant