# Dynamics - the physics of tobogganing and forces at an angle

## Homework Statement

A child is tobogganing down a hillside. The child and the toboggan together have a mass of 50.0kg. The slope is at an angle of 30.0° to the horizontal.
Assume that the positive y-direction is pointing in the direction of the normal force. Assume that the positive x-direction is down the incline.

Find the acceleration of the child

a) in the case where there is no friction

b) if the coefficient of friction is 0.15

## The Attempt at a Solution

a) Fn + FgI = mgcosθ
= (50)(9.8)cos 30°
= 424N
Fnet = FgII = mgsinθ
= (50)(9.8)sin 30°
= 245 N (+ x-direction)
Fnet = ma
a = Fnet/m
= 245/50
= +4.9m/s/s
Therefor the child's acceleration in the case where there is no friction is 4.9 m/s/s.

b) FgI = 424 (- y-direction)
FgII = 245 (+ x-direction)

Fnet = FgII (+ x-direction) + Fk (- x-direction)

Fn = FgI = 424N (down)

Fk = μkFn
= (0.15)(424)
= 63.6N (- x-direction)

Fnet = FgII + Fk
= (+245)+(-63.6)
= +181.4N
Fnet =ma
a = Fnet/m
= 181.4/50
= +3.63m/s/s

Therefore the child's acceleration is +3.63 m/s/s if the coefficient of friction is 0.15.

Dick
Science Advisor
Homework Helper

## Homework Statement

A child is tobogganing down a hillside. The child and the toboggan together have a mass of 50.0kg. The slope is at an angle of 30.0° to the horizontal.
Assume that the positive y-direction is pointing in the direction of the normal force. Assume that the positive x-direction is down the incline.

Find the acceleration of the child

a) in the case where there is no friction

b) if the coefficient of friction is 0.15

## The Attempt at a Solution

a) Fn + FgI = mgcosθ
= (50)(9.8)cos 30°
= 424N
Fnet = FgII = mgsinθ
= (50)(9.8)sin 30°
= 245 N (+ x-direction)
Fnet = ma
a = Fnet/m
= 245/50
= +4.9m/s/s
Therefor the child's acceleration in the case where there is no friction is 4.9 m/s/s.

b) FgI = 424 (- y-direction)
FgII = 245 (+ x-direction)

Fnet = FgII (+ x-direction) + Fk (- x-direction)

Fn = FgI = 424N (down)

Fk = μkFn
= (0.15)(424)
= 63.6N (- x-direction)

Fnet = FgII + Fk
= (+245)+(-63.6)
= +181.4N
Fnet =ma
a = Fnet/m
= 181.4/50
= +3.63m/s/s

Therefore the child's acceleration is +3.63 m/s/s if the coefficient of friction is 0.15.

I think your answers are right. I'm having a hard time puzzling through your designations for the various forces though. Could you explain them?

yeah I wasn't sure how to write them on the computer, basically I am following my book.
Fg - force of gravity
FgI - Fg(cosθ)
FgII = Fg(sinθ)
Fn - normal force
Fnet - net force
Fk - force of kinetic friction
m - mass
a - acceleration
g - gravity
μk - coefficient of kinetic friction

please let me know if I missed any or if you still have any questions. Thanks