Calculating force needed to push sled down incline

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Homework Help Overview

The original poster attempts to calculate the force required to push a sled down a 6-degree incline, given its mass, desired ending speed, distance traveled, and the coefficient of friction. The problem involves concepts from dynamics and kinematics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of force equations and the application of Newton's laws. There are attempts to clarify the signs used in calculations, particularly regarding gravitational acceleration and friction. Some participants question the consistency of sign conventions applied in the original poster's calculations.

Discussion Status

There is ongoing dialogue about the correct application of signs in the equations. Participants have offered guidance on maintaining consistent sign conventions and have pointed out potential errors in the original poster's reasoning. The discussion is focused on refining the approach rather than reaching a definitive solution.

Contextual Notes

Participants note the importance of sign conventions in physics problems, particularly when dealing with forces acting in different directions. There is an acknowledgment of varying interpretations of gravitational acceleration in different contexts.

jmm5180
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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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jmm5180 said:
Fa= 22 ( (0.2)(-9.8)(cos6) - (-9.8sin6) + 1.85 ))
Careful with signs. g is just a positive constant.
 
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.
 
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?
 
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.)
 

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