A toboggan and rider of

total mass 90 kg travel

down along the (smooth)

slope defined by the

equation y = 0.08x2.

At the instant x = 10,

the toboggans speed is 5 m/s.

At this point, determine the rate of increase in

speed and the normal force which the slope

exerts on the toboggan. Neglect the size of the

toboggan and rider for the Calculation.

Quiz

Solution FBD y = 0.08x2

Therefore,

The slope angle at x = 10 m is given by

tan =

= 0.16(10)

= 57.99

16.0

16.0

2

2

dx

yd

xdx

dy

xdx

dy

x

16.010

And the radius of curvature at x = 10 m is

m98.41

16.0

16.011

10

2

32

2

2

2

32

x

x

dx

yd

dx

dy

y

x

Solution (continued) = 57.99 and = 41.98 m

Ft = mat 90(9.81) sin 57.99 = 90at at = 8.32 m/s

2

Fn = man -90(9.81) cos 57.99 + N = 90

N = 522 N

98.41

52