Calculating Normal Force on a Snow Sled with Newton's Laws

[roam]

(a) A snow sled with a child secured safely to it has a total mass of 56.0 kg. It is lowered at a constant speed of 1.1 m s−1 down a slope of angle 35.0 ° with respect to the horizontal (as shown above) for a distance d = 18.0 m. The coefficient of kinetic friction between the sled and the snow is 0.13.

Note: g = 9.8 m s–2. Air resistance is negligible at these speeds.

Find the magnitude of the reactive force, N, on the sled.




The correct answer should be 450.0 but my answer is different. I don't know how to solve this question but here's my attemp:



ΣFx = 1.1 cos 35.0° = 0.90
ΣFy = 1.1 sin 35.0 = 0.63



a = \frac{\sum F}{m}

ax = 0.90 ÷ 56.0 = 0.01
ay = 0.63 ÷ 56.0 = 0.01

Now to find the magnitude:

\sqrt{0.01^2 + 0.01^2} = 0.01 "?"

≠ 450.0 N

Thanks.

 

[Nabeshin]

The coefficient of kinetic friction between the sled and the snow is 0.13.

You failed to incorporate this into your solution. Taking this into consideration should help ^^

 

[roam]

You failed to incorporate this into your solution. Taking this into consideration should help ^^

How can I incorporate that in my solution?

ƒk = μkn
ƒ = 0.13 \times n

So, what should I use for the n, the magnitude of the normal force?

 

[Nabeshin]

Well, you had no problem resolving the gravity force into it's down-the-hill component. Seeing as the object seems not to be accelerating in the new "y" direction, what can we say about the forces there? This should help you solve for normal force.