Question

(a) If half of the weight of a small 1.00×103kg utility truck is supported by its two drive wheels, what is the magnitude of the maximum acceleration it can achieve on dry concrete? (b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate? (c) Solve both problems assuming the truck has four-wheel drive.

Answer 1

Answer: $4.9 m/s^{2}$

Explanation:

Start by finding the weight supported by the wheels, as we can then relate that to the total frictional force through the friction coefficient. This is given by:

W=mg

Substitute 1.00×$10^{3}$, or 1,000, for m and 9.81 m/$s^{2}$ for g. This gives us

W=(1,000)(9.81)

W=9,810 N

Since half the weight is supported by the wheels in question, we divide this by two, and the weight we will use is 4,905 N.

The frictional force is given by:

F=μW

Where μ is the friction coefficient and W is the weight, which we just calculated. The friction coefficient for rubber on dry concrete is given in Table 6.1 as (1.0). Substitute that for μ and 4,905 N for weight and we get the total frictional force as 4,905 N. Now we can find the acceleration by rearranging F=ma to get a=$\frac{F}{m}$. Substitute our frictional force 4,905 N for F and our total mass 1,000 kg for m to get 4.9 m/$s^{2}$.