Calculating force needed to push sled down incline

In summary: Just be consistent.In summary, the force required to push a 22kg sled down a 6 degree hill and have an ending speed of 60km/hr after traveling 75m when the coefficient of friction is 0.20 is 20.36 N. The correct approach is to use g as a positive constant and assign signs consistently with the chosen sign convention.
  • #1
jmm5180
22
0

Homework Statement


Calculate the force required to push a 22kg sled down a 6 degree hill and have an ending speed of 60km/hr after traveling 75m when the coefficient of friction is 0.20.

Homework Equations


Fnet= ma
(V2)^2 = (V1)^2 + 2a(d2-d1)
Ff= uN
W= mg

The Attempt at a Solution


I'll use @ for theta, and u for mu. Fa= force applied to the sled, Ff=force of friction

With N lying on the positive Y axis and Weight in the x direction lying on the positive x axis, I set Wy=wcos@ and Wx= wsin@, so N=wcos@.

I tried summation of forces:
xFnet = Wx + Fa - Ff = ma
Fa= Ff - Wx + ma
Fa = uN - wsin@ + ma
Fa = u(wcos@) - wsin@ + ma
Fa = umgcos@ - mgsin@ + ma
Fa = m(ugcos@ - gsin@ + a)

I then tried to solve for a using (V2)^2 = (V1)^2 + 2a(d2-d1)
Since the starting velocity was 0 and the starting distance is 0,
A = (V2)^2 / (2d2)

I converted 60 km/hr to m/s, to get 16.66 m/s.
A= (16.66^2)/ 2(75) = 1.85 m/s^2

Plugging that in...

Fa= 22 ( (0.2)(-9.8)(cos6) - (-9.8sin6) + 1.85 ))
I got Fa = 20.36 N, though that isn't correct. I'm not sure what I did wrong.
 
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  • #2
jmm5180 said:
Fa= 22 ( (0.2)(-9.8)(cos6) - (-9.8sin6) + 1.85 ))
Careful with signs. g is just a positive constant.
 
  • #3
Check the signs of the forces. Looks like you have the required net force (ma) the same sign as the friction force (uN)?

May not be the only problem but that looks wrong to me.
 
  • #4
Doc Al said:
Careful with signs. g is just a positive constant.

Thank you. I've used -9.8 m/s^2 before in questions and have gotten the correct answer. How do I know when to use the positive/negative version?
 
  • #5
jmm5180 said:
Thank you. I've used -9.8 m/s^2 before in questions and have gotten the correct answer. How do I know when to use the positive/negative version?
I would always use g as a positive constant and add signs as needed. For example, using down as negative, the acceleration of a falling object is -g = -9.8m/s^2.

When you set up your force equations, always assign your signs consistent with whatever sign convention you are using. (For example, taking down the incline as positive would make the weight component, the applied force, and the acceleration positive and friction negative.)
 

1. What factors affect the amount of force needed to push a sled down an incline?

The amount of force needed to push a sled down an incline is affected by several factors, including the mass of the sled, the angle of the incline, and the coefficient of friction between the sled and the incline surface.

2. How do I calculate the force needed to push a sled down an incline?

The force needed to push a sled down an incline can be calculated using the formula F = mgsinθ, where F is the force, m is the mass of the sled, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the incline.

3. What is the role of friction in calculating the force needed to push a sled down an incline?

Friction plays a significant role in calculating the force needed to push a sled down an incline. It is the force that opposes motion and must be overcome by the pushing force. The coefficient of friction between the sled and the incline surface is used to calculate the frictional force.

4. How does the angle of the incline affect the force needed to push a sled down?

The steeper the incline, the greater the force needed to push a sled down. This is because as the angle of the incline increases, the component of the force acting against the sled's motion also increases, making it harder to push the sled.

5. Is the force needed to push a sled down an incline the same as the force needed to pull the sled up the incline?

No, the force needed to push a sled down an incline is not the same as the force needed to pull the sled up the incline. When pulling a sled up an incline, you are working against gravity, so the force needed will be greater than when pushing the sled down the incline.

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