defined by the equation y=0.08.x². At the instant x - 10 m, the toboggan's 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. -5m/s/ y=0.08.² 10 m

defined by the equation y=0.08.x². At the instant x - 10 m, the toboggan's 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. -5m/s/ y=0.08.² 10 m

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Laws of Equilibrium

A body is said to be in equilibrium if the system of forces acting on the body tends to keep the body at rest, in other words, there is no translation motion or net torque on the body. Under such conditions, the equilibrium conditions are to be applied to analyze the condition. The branch of mechanics that deals with this condition is known as 'statics'. Under equilibrium, there are no net forces on the body.

Question

3. 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 m, the toboggan's 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.
v=5m/s/
y=0.08.²
10 m

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