Question

A 1500 kg car sits on a 3.5° inclined hill. Find the force of friction required to keep it from

sliding down the hill. The coefficient of static friction is μ=0.45

Answer 1

The force of friction required to keep the car from sliding down the hill is 898.054 newtons and since
`f<f_(max)`, the car will not slide down the hill.

Step-by-step explanation:

The free body diagram is included below and now we prepare each equation of equilibrium for respective orthogonal axis:

`\Sigma F_(x') = f-m\cdot g \cdot \sin \theta = 0` (1)

`\Sigma F_(y') = N-m\cdot g \cdot \cos \theta = 0` (2)

Where:

`f` - Static friction force, measured in newtons.

`N` - Normal force from the inclined hill to the car, measured in newtons.

`m` - Mass of the car, measured in newtons.

`g` - Gravitational constant, measured in meters per square second.

`\theta` - Angle of inclination of the hill, measured in sexagesimal degrees.

If we know that
`m = 1500\,kg`,
`g = 9.807\,(m)/(s^(2))` and
`\theta = 3.5^(\circ)`, then the friction force required to keep the car from sliding down the hill is:

`f = m\cdot g \cdot \sin \theta`

`f = (1500\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot \sin 3.5^(\circ)`

`f = 898.054\,N`

The force of friction required to keep the car from sliding down the hill is 898.054 newtons.

If the car does not slide down the hill, then the following condition must be observed:

`f\le \mu_(s)\cdot m\cdot g \cdot \cos \theta` (2)

Where
`\mu_(s)` is the static coefficient of friction, dimensionless.

If we know that
`\mu_(s) = 0.45`,
`m = 1500\,kg`,
`g = 9.807\,(m)/(s^(2))` and
`\theta = 3.5^(\circ)`, then the maximum allowable static friction force is:

`f_(max) = \mu_(s)\cdot m \cdot g \cdot \cos \theta`

`f_(max) = (0.45)\cdot (1500\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot \cos 3.5^(\circ)`

`f_(max) = 6607.378\,N`

Since
`f<f_(max)`, the car will not slide down the hill.