Question

A cyclist is moving up a slope that is at an angle of 19 to the horizontal. The mass of the cyclist and the bicycle is 85 kg. What is the component of the weight of the cyclist and bicycle parallel to the slope and what is the normal reaction force on the bicycle from the slope?

Answer 1

Answer:

Explanation:

Given

Slope of inclination $\theta =19^{\circ}$

Mass of cyclist and bicycle is $m=85\ kg$

When cyclist is going up then there is two components of its weight , one is parallel to inclination and other is perpendicular to inclination

Weight of person is mg

it can be resolved in $mg\cos \theta$ and  $mg\sin \theta$

$mg\cos \theta$ is perpendicular to the inclination

and $mg\sin \theta$ is parallel to inclination as shown in diagram

Parallel component $=mg\sin \theta=85\times 9.8\times \sin 19=271.2\ N$

Normal reaction $N=mg\cos \theta =85\times 9.8\times \cos 19=787.61\ N$

Answer 2

The normal reaction on the bicycle by the inclined slope is 787.62 N.

The parallel on the bicycle along the inclined slope is 271.2 N.

The given parameters;

• angle of the slope, = 19
• mass of the bicycle, m = 85 kg

The normal reaction on the bicycle by the inclined slope is calculated as; follows;

$F_n = W cos \theta\\\\F_n = mg \ cos \theta \\\\F_n = 85 \times 9.8 \times cos \ (19)\\\\F_n = 787.62 \ N$

The parallel on the bicycle along the inclined slope is calculated as

$F_x = Wsin\theta\\\\F_x = mg \ sin \ \theta \\\\F_x = 85 \times 9.8 \times sin(19) \\\\F_ x = 271 .2 \ N$

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