Question

The coefficient of static friction between the tires of a car and a dry road is \(0.62\). The mass of the car is \(1500 \mathrm{~kg}\). What maximum braking force is obtainable \((a)\) on a level road and \((b)\) on an \(8.6^{\circ}\) downgrade?

Short answer

The maximum braking force obtainable on a level road is approximately \(9114 \mathrm{~N}\) and on an 8.6 degree downgrade it is approximately \(8938.54 \mathrm{~N}\).

Step-by-step solution

Calculate the normal force on a level road

The normal force is the force exerted by a surface to support the weight of an object resting on it. On a level road, normal force (\(F_n\)) can be calculated using the formula: \(F_n = m*g\) where \(m\) is the mass of the car and \(g\) is the gravitational acceleration. Here, \(m=1500 \mathrm{~kg}\) and \(g=9.8 \mathrm{~m/s^2}\). Hence, \(F_n=1500*9.8 = 14700 \mathrm{~N}\).

Calculate the maximum braking force on a level road

The maximum static friction force (\(F_{\max}\)) is given by \(F_{\max} = μ*F_n\), where \(μ\) is the coefficient of static friction. Here, \(μ=0.62\) and \(F_n=14700 \mathrm{~N}\). So, \(F_{\max} = 0.62*14700 = 9114 \mathrm{~N}\).

Calculate the normal force on an 8.6-degree downgrade

On a downgrade, the normal force changes due to the slope of the road. Normal force (\(F'_n\)) on a slope can be computed using the formula: \(F'_n = m*g*cos(θ)\), where \(θ\) is the downgrade angle. Here, \(θ=8.6^{\circ}\) (converted to radians = 0.15 rad). So, \(F'_n = 1500*9.8*cos(0.15) = 14417 \mathrm{~N}\).

Calculate the maximum braking force on an 8.6-degree downgrade

We apply the same formula we used in step 2 to compute the maximum static friction force on a downgrade (\(F'_max\)), using the normal force on a downgrade (\(F'_n\)). Hence, \(F'_max = μ*F'_n = 0.62*14417 = 8938.54 \mathrm{~N}\).

Key concepts

Static Friction

Imagine you're in a car, pressing the brakes to avoid hitting an obstacle. The immediate force that resists the sliding of the car tires against the road is called static friction. This force plays a crucial role in determining how quickly a vehicle can come to a stop without slipping. Static friction is at its strongest just before the object starts moving; this is the 'maximum static friction force'.

For a car, this force is crucial when braking as it determines the maximum force that can be applied before the tires start to skid. In our exercise, we found this force by multiplying the coefficient of static friction (a unitless value representing the friction between two surfaces) by the normal force. The coefficient of static friction is unique to the materials in contact – in this case, the car tires and the road. Here, it was given as 0.62. This means that for every 1 Newton of normal force, the maximum static friction force that can be applied without causing movement is 0.62 Newtons.

Normal Force

When we talk about the normal force, we're referring to the support force exerted by a surface when an object is resting on it. It's the upward force that counteracts gravity, ensuring that objects like a car don’t just sink through the pavement. On a flat surface, the normal force (\( F_n \) is typically equal to the force due to gravity on the object, assuming there's no vertical acceleration.

In our example, the car's mass is 1500 kg, and when multiplied by the gravitational acceleration (9.8 m/s²), we calculate a normal force of 14700 N on a level road. This normal force changes if the surface is inclined, like a road on a hill. For a slope, the normal force is always perpendicular to the surface, and thus we see it affected by the cosine of the angle of the slope. For angles that are not steep, the normal force will only be slightly less than on a flat surface. That is why on an 8.6-degree downgrade, the normal force decreases slightly from 14700 N to 14417 N.

Gravitational Acceleration

Gravitational acceleration is the consistent pull that gravity exerts on an object, directed towards the center of the Earth. It is denoted by the symbol 'g' and approximately equals 9.8 m/s² at the Earth's surface. This acceleration is the reason objects fall towards the ground when dropped and why a car's weight presses down on the road.

When we calculated the normal force, we used the gravitational acceleration value in our formula. It's crucial for understanding how forces like the normal force are calculated. Since gravitational acceleration is constant near the Earth's surface, it serves as a reliable value for many physics problems involving weight and mass. However, it's worth noting that this value can be slightly different depending on altitude and location due to the Earth's not perfectly uniform composition.