Question

A lorry, of mass 38000 kg, starts up a hill of gradient 1 in 12. The constant acceleration is 0.06 m/s2 and resistance to motion is 1200 N (not gravitational force). What is the tractive force in Newtons exerted by the lorry’s driving wheels?

Answer 1

Answer on Question#53383 - Physics - Mechanics - Kinematics - Dynamics

A lorry, of mass $m = 38000 \, \mathrm{kg}$, starts up a hill of gradient 1 in 12. The constant acceleration is $a = 0.06 \frac{\mathrm{m}}{\mathrm{s}^2}$ and resistance to motion is $F_r = 1200 \, \mathrm{N}$ (not gravitational force). What is the tractive force $F$ in Newton's exerted by the lorry's driving wheels?

Solution:

According to the Newton's second law we obtain

$$m a = F - \left(F _ {r} + m g \sin \phi\right),$$

where $g = 10\frac{\mathrm{m}}{\mathrm{s}^2}$ - acceleration due to gravity, and $\phi$ - is the angle of the incline. It's given that $\tan \phi = \frac{1}{12}$, or equivalently $\cot \phi = 12$. Since

$$\sin \phi = \frac {1}{\sqrt {1 + \cot^ {2} \phi}},$$

we obtain

$$m a = F - \left(F _ {r} + \frac {m g}{\sqrt {1 + \cot^ {2} \phi}}\right)$$ $$\begin{array}{l} F = m a + F _ {r} + \frac {m g}{\sqrt {1 + \cot^ {2} \phi}} = 3 8 0 0 0 \mathrm {k g} \cdot 0. 0 6 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} + 1 2 0 0 \mathrm {N} + \frac {3 8 0 0 0 \mathrm {k g} \cdot 1 0 \frac {\mathrm {m}}{\mathrm {s} ^ {2}}}{\sqrt {1 + 1 4 4}} = \\ = 3 5 0 3 7 \mathrm {N} \\ \end{array}$$

Answer: 35037 N.

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