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

AI Thread Summary
A horse with a mass of 509 kg pulls a sleigh weighing 255 kg, both accelerating at 0.5 m/s², with a friction coefficient of 0.150. The calculated force of friction acting on the sleigh is 749.7 N. To find the total force exerted by the horse, the equation Ft - Ff = (m + M)a is used, leading to a required force of 3503.2 N. Discussions highlight that the pulling force must overcome friction to initiate movement, and theoretically, without friction, even a gentle push could suffice to move the sleigh. The conversation emphasizes the importance of understanding net forces in the horizontal direction for accurate calculations.
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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