How to find Coefficient of Friction w/ Fa and mass?

AI Thread Summary
To find the coefficient of kinetic friction for a sled being pulled by poodles, the net force is calculated to be zero since the sled moves at a constant velocity. The applied force is 66N, and the normal force is determined using Fn = mg, resulting in 137.2N. The frictional force is then calculated using the equation Ff = u * Fn, leading to the coefficient of friction u being approximately 0.48. The calculations are confirmed to be correct, indicating a solid understanding of the problem. This method effectively demonstrates how to apply physics equations to find the coefficient of friction.
HomiesontheRise
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Homework Statement


A team of six poodles are pulling a 14.0 kg sled at 8.0 m/s

IF the dogs are applying a 66N force, find the coefficient of kinetic friction

Homework Equations


Ff=u*Fn
Fnet=ma
Fnet=(-Ff)+Fa

The Attempt at a Solution


I honestly have no clue where to start for this question.
 
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HomiesontheRise said:

Homework Statement


A team of six poodles are pulling a 14.0 kg sled at 8.0 m/s

IF the dogs are applying a 66N force, find the coefficient of kinetic friction

Homework Equations


Ff=u*Fn
Fnet=ma
Fnet=(-Ff)+Fa

The Attempt at a Solution


I honestly have no clue where to start for this question.
HomiesontheRise said:

Homework Statement


A team of six poodles are pulling a 14.0 kg sled at 8.0 m/s

IF the dogs are applying a 66N force, find the coefficient of kinetic friction

Homework Equations


Ff=u*Fn
Fnet=ma
Fnet=(-Ff)+Fa

The Attempt at a Solution


I honestly have no clue where to start for this question.
Well you have correctly listed the relevant equations, now first identify what the acceleration is and solve. Read the problem carefully.
 
So I started by finding the Net Force which was 0 due to the fact that there is no acceleration (velocity is constant).
Then found Force of Friction using Fnet and Fa with the formula Fnet=Fa+(-Ff).
Then found Fn using Fn=mg.
Finally, I user Ff=Fn*u to find u

MY WORK
1.
Fnet=0 Fnet-Fa=-Ff 0-66=-Ff -66=-Ff
2.
Fn=mg Fn=14kg*9.8 Fn=137.2
3.
Ff=u*Fn u=Ff/Fn 66/137.2=u 0.48=u
 
Please tell me if you see any errors in my calculations
 
PhanthomJay said:
Well you have correctly listed the relevant equations, now first identify what the acceleration is and solve. Read the problem carefully.
Does this look right?
1.
Fnet=0 Fnet-Fa=-Ff 0-66=-Ff -66=-Ff
2.
Fn=mg Fn=14kg*9.8 Fn=137.2
3.
Ff=u*Fn u=Ff/Fn 66/137.2=u 0.48=u
 
Look
HomiesontheRise said:
Does this look right?
1.
Fnet=0 Fnet-Fa=-Ff 0-66=-Ff -66=-Ff
2.
Fn=mg Fn=14kg*9.8 Fn=137.2
3.
Ff=u*Fn u=Ff/Fn 66/137.2=u 0.48=u
Looks good!
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
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