Question

Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the static coefficient of friction is involved—that is, the tires are not allowed to slip during the deceleration.

Answer 1

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration  of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²

Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

$F_{net}$ = - $W_x$ - $f_s$

The $W_x$ is a x-component of force due to gravity (W) and, in this case, is given by: $W_x$ = W.sin(14)

W is described as: W = m.g

Force due to friction ($f_s$) is given by: $f_s$ = μs.N

N is the normal force and, in the system, is equivalent of $W_y$, so:

$W_y$ = m.g.cos(14)

Therefore, the formula will be:

$F_{net}$ = - $W_x$ - $f_s$

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²