How Much Force Does a Horse Need to Pull a Sleigh?

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Homework Help Overview

The discussion revolves around calculating the force required for a horse to pull a sleigh, considering the masses involved, the acceleration, and the coefficient of friction between the sleigh and the snow. Participants are analyzing the forces acting on the system, including friction and net force calculations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to calculate the frictional force and the total force exerted by the horse. There are discussions about the correct interpretation of the forces in the x-direction and the impact of the sleigh's weight on the calculations. Some participants question the inclusion of certain terms in the equations and explore the implications of different friction values.

Discussion Status

Multiple interpretations of the problem are being explored, with some participants providing calculations and others questioning the validity of those calculations. There is no explicit consensus, but guidance is being offered regarding the setup of the equations and the roles of different forces.

Contextual Notes

Participants are navigating potential confusion regarding the problem setup, particularly concerning the frictional force and the weight of the sleigh. There is an acknowledgment of the theoretical aspects of friction and movement in the context of the problem.

Muteb
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A hors og mass 509 KG Pulls a sleigh of mass 255 KG and both horse and sleigh accelerate 0.5 m/s^2. the coefficient of friction for the sleigh is 0.150 as it moves over the snow. Find the horizontal force that the horse must exert on the sleigh. Remember sleighs have two identical blades on the snow

Ff= 0.150*9.8*255*2= 749.7N
Ft-(255*9.8)-(749.7)=509*0.5
Ft-3248.7=254.5
Ft=254.5+3248.7
Ft= 3503.2N
 
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Muteb said:
A hors og mass 509 KG Pulls a sleigh of mass 255 KG and both horse and sleigh accelerate 0.5 m/s^2. the coefficient of friction for the sleigh is 0.150 as it moves over the snow. Find the horizontal force that the horse must exert on the sleigh. Remember sleighs have two identical blades on the snow

Ff= 0.150*9.8*255*2= 749.7N
Ff = uN, where N=the weight of the sleigh = 255*9.8N. That '2' doesn't belong there; the problem was trying to throw you off. Ff = 375N , and it acts in the negative x direction.
Ft-(255*9.8)-(749.7)=509*0.5
Ft-3248.7=254.5
Ft=254.5+3248.7
Ft= 3503.2N
I assume by Ft you mean the force of the horse on the sleigh. Ft is in the x direction. For starters, that sleigh weight, 255*9.8, which acts in the vertical direction, doesn't belong there. You're looking for net forces in the x direction. Look at the horse and sleigh together. In the x direction, F_net = (m +M)a, where (m +M) is the combined weight of sleigh and horse, and F_net , the forces pulling on the sleigh in the x direction, is (Ft -Ff). So it's Ft -Ff = (m + M)a. Canyou sove for Ft?
 


I think that

Ft=375 + 382
Ft=757N
But do you think the force (757N) can move the sleigh ( 2499N)?
 


Muteb said:
I think that

Ft=375 + 382
Ft=757N
But do you think the force (757N) can move the sleigh ( 2499N)?
What do you think? Ever been to an airport and seen a rinky-dink tow truck pull a 1,000,000N 707 jetliner?
Why is this so?
 


Ft=375 + 382
Ft=757N
i that right
 


Muteb said:
Ft=375 + 382
Ft=757N
i that right
looks ok, but you didn't answer my question.
 


Thanks for your help

I don't know about it but could you tell me
 


Muteb said:
Thanks for your help

I don't know about it but could you tell me
The pulling force need only overcome friction to move it. If there were ideally an theoretically no friction between sleigh and snow, you wouldn't need to pull it with a horse to move it, you could move it with a gentle shove.
 

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