Dynamics - the physics of tobogganing and forces at an angle

  • Thread starter L_0611
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
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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?
 
  • #3
24
0
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
 

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