Question

A cyclist starts from rest down a hill, of gradient 1 in 30. The gearing allows a constant tractive force of 180 N. The resistance to motion is 30 N. The total mass of cycle and cyclist is 85 kg. Calculate the constant acceleration in m/s2

Answer 1

Answer on Question #44614, Physics, Mechanics | Kinematics | Dynamics

A cyclist starts from rest down a hill, of gradient 1 in 30. The gearing allows a constant tractive force of $180\mathrm{N}$ . The resistance to motion is $30\mathrm{N}$ . The total mass of cycle and cyclist is $85\mathrm{kg}$ . Calculate the constant acceleration in $\mathrm{m/s}^2$ .

Solution:

Given:

$m = 85\mathrm{kg}$

$F = 180\mathrm{N}$

$F_{fr} = f = 30\mathrm{N},$

$a = ?$

If the angle is expressed as a ratio (1 in n) then:

$$\theta = \tan^ {- 1} \left(\frac {1}{n}\right)$$

Thus,

$$\theta = \tan^ {- 1} \left(\frac {1}{3 0}\right) = 1. 9 0 9 {}^ {\circ}$$

The equation of motion is

$$m a = F + m g \sin \theta - F _ {f r}$$

where $g$ is the gravity acceleration constant $(9.81\mathrm{m} / \mathrm{s}^2)$ .

The acceleration is

$$a = \frac {F}{m} + g \sin \theta - \frac {F _ {f r}}{m}$$

Thus,

$$a = \frac {1 8 0}{8 5} + 9. 8 1 \cdot \sin 1. 9 0 9 {}^ {\circ} - \frac {3 0}{8 5} = 2. 0 9 1 = 2. 1 \mathrm {m / s ^ {2}}$$

Answer. $a = 2.1 \, \text{m/s}^2$ .

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