Determine normal force and friction of a toboggan with children

[thomasrules]

I've tried it but I got stuck at the part to find the coefficient of friction.

A parent pulls a toboggan with three children at a constant velocity for 38m along a horizontal trail. The total mass of the children and the toboggan is 66kg. The force the parent exerts is 58N[18 above the horizontal]

a)Determine the magnitude of the norml force and the coefficient of kinetic friction.

 

[Nylex]

What have you done so far? What does constant velocity tell you?

 

[Andrew Mason]

I've tried it but I got stuck at the part to find the coefficient of friction.

A parent pulls a toboggan with three children at a constant velocity for 38m along a horizontal trail. The total mass of the children and the toboggan is 66kg. The force the parent exerts is 58N[18 above the horizontal]

a)Determine the magnitude of the norml force and the coefficient of kinetic friction.

The horizontal component of the pulling force balances the friction force.

So:
$$F_{friction} = \mu_kF_N = 58Ncos(18)$$

All you have to do is work out the normal force. Be careful. There are two components to the parent's pulling force. The vertical component reduces the normal force.

AM

 

[thomasrules]

Is FRICTION EQUAL TO WORK Facos18 because net f=0?

and the normal force would be fg=fn=mg?

 

[Andrew Mason]

Is FRICTION EQUAL TO WORK Facos18 because net f=0?

and the normal force would be fg=fn=mg?

No. See my previous post.

(1)$$F_N = mg - F_{pull}sin(\theta)$$

(2)$$\mu_kF_N = F_{pull}cos(\theta)$$

Substitute $F_N$ from (1) into (2) to get the coefficient of friction.

AM

 

[thomasrules]

but why wouldn't fn=fg...doesn't make sense

 

[Andrew Mason]

but why wouldn't fn=fg...doesn't make sense

There is an upward lift force due to the angle at which the parent is pulling the toboggan. This lift reduces the downward mg force.

AM