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

In summary, the horizontal force that the horse must exert on the sleigh is 757N. This force is enough to overcome the friction and move the sleigh with an acceleration of 0.5 m/s^2. The coefficient of friction for the sleigh is 0.150 and its weight is 255*9.8N. However, the pulling force does not need to exceed the weight of the sleigh, as it only needs to overcome the friction for the sleigh to move.
  • #1
Muteb
26
0
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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  • #2


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?
 
  • #3


I think that

Ft=375 + 382
Ft=757N
But do you think the force (757N) can move the sleigh ( 2499N)?
 
  • #4


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?
 
  • #5


Ft=375 + 382
Ft=757N
i that right
 
  • #6


Muteb said:
Ft=375 + 382
Ft=757N
i that right
looks ok, but you didn't answer my question.
 
  • #7


Thanks for your help

I don't know about it but could you tell me
 
  • #8


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.
 

1. How do you calculate the horizontal force?

The horizontal force can be calculated using Newton's Second Law, which states that force equals mass times acceleration. In the case of horizontal force, the acceleration is the change in velocity over time in the horizontal direction.

2. What is the difference between horizontal force and vertical force?

Horizontal force refers to the force acting in the horizontal direction, while vertical force is the force acting in the vertical direction. They are perpendicular to each other and can have different magnitudes and directions.

3. How is horizontal force used in real-world applications?

Horizontal force is used in various real-world applications, such as pushing or pulling objects, moving vehicles, and designing structures to withstand lateral forces like wind or earthquakes.

4. What factors can affect the magnitude of horizontal force?

The magnitude of horizontal force can be affected by the mass and acceleration of the object, as well as external factors such as friction, air resistance, and the angle at which the force is applied.

5. How do you find the horizontal force in a vector diagram?

In a vector diagram, the horizontal force is represented by the horizontal component of the force vector. It can be found by using trigonometric functions to calculate the horizontal and vertical components of the vector.

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