# Homework Help: Find the horizontal force

1. Nov 23, 2008

### Muteb

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

2. Nov 24, 2008

### PhanthomJay

Re: Help

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.
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. Nov 24, 2008

### Muteb

Re: Help

I think that

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

4. Nov 24, 2008

### PhanthomJay

Re: Help

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. Nov 24, 2008

### Muteb

Re: Help

Ft=375 + 382
Ft=757N
i that right

6. Nov 24, 2008

### PhanthomJay

Re: Help

looks ok, but you didn't answer my question.

7. Nov 24, 2008

Re: Help