Problems on Friction

Effects of Friction on Acceleration

The following problems prove that friction, which usually works in the opposite direction to motion, affects acceleration. As per Newton’s second law of motion, friction cancels part of the force, leading to a smaller acceleration.

Question 1: A locomotive train consists of two 8.00 × 10⁵ kg engines and 45 cars with average masses of 5.50 × 10⁵ kg.

1. What force must each engine exert backward on the track to accelerate the train at a rate of 5.00 × 10^-2 m/s² if the force of friction is 7.50 × 10⁵N, assuming the engines exert identical forces?

First, let's gather all the relevant information and calculations.

Given that:

Mass of each engine =8.00 × 10⁵ kg

Number of engines = 2

Mass of each car = 5.50 × 10⁵ kg

Number of cars = 45

Total mass of engines: 2× (8.00 × 10⁵ )= 1.60 × 10⁶kg

Total mass of cars: 45 × (5.50 × 10⁵ )= 2.475 × 10⁷ kg

Total mass of the train=total mass of engines + total mass of cars

The total mass of the train: 1.60 × 10⁶kg + 2.475 × 10⁷ kg

Calculating the total force required to accelerate the train:

Acceleration (a)= 5.00 × 10^-2 m/s²

Therefore, using Newton’s second law:

F _total= m × a

F _total = (2.635× 10⁷) × (5.00× 10^-2)

F _total = 1.3175 × 10⁶ N

Calculating the total force exerted by the engines, accounting for friction:

F _engine =F _total + F _friction

F _engine = 1.3175 ×10⁶N + 7.50 × 10⁵N= 2.0675 × 10⁶N

Since there are two engines exerting equal force, the force each engine must exert is:

F _{engine = (}F _engine)/2 ₌ (2.0675 x 10⁶N)/2= 1.03375 × 10⁶ N

1. What is the magnitude of the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?

To find the coupling force between the 37^th and 38^th cars, we need to determine the mass of all cars behind the 37^th car (including the 38^th car). The cars being pulled by the coupling are cars numbered 38 through 45. Therefore, this is a total of 8 cars:

45- 37= 8 cars

Mass of one car = 5.50 × 10⁵ kg

The total mass of cars being pulled by the coupling:

m= 8 × 5.50 × 10⁵ kg

m= 4.40 × 10⁶ kg

Calculating the friction force acting on the cars and the engines, assuming that friction is evenly distributed:

friction force per car and engine= F _friction / (45+2) = (7.50 ×10⁵N)/47

=1.5957 × 10⁴ N (since there are 45 cars and 2 engines)

Total friction force affecting 8 cars:

Friction force on 8 cars = 8 × (1.5957 × 10⁴ N) = 1.2766 ×10⁵N

Lastly, calculating the total force required to accelerate 8 cars, including friction:

The total mass of 8 cars = 4.40 ×10⁶ kg

Force needed for acceleration = 4.40 ×10⁶ ×5.00× 10^-2 = 2.20 × 10⁵N

Total force including friction = 2.20 × 10⁵ +1.2766 ×10⁵N= 3.4766 × 10⁵N

The force’s magnitude in the coupling between the 37^th and 38^th cars after accounting for eventually distributed friction is 3.4766 × 10⁵N

Question 2: Show that the acceleration of any object down an incline where friction behaves simply (that is, where f_k=μ_kN is a=g (sinθ - μ_kcosθ).  

To denote that acceleration a of an object down an incline where kinetic friction f_k=μ_kN is given by a=g (sinθ - μ_kcosθ), it is imperative to analyze the force acting on the object and derive the expression for the acceleration.

First, let us identify the forces:

• Gravitational force (mg) acting vertically downward.
• Normal force (N), which is perpendicular to the surface of the incline.
• Kinetic friction force f_k=μ_kN acting parallel to the surface of the incline and opposing the motion.

Now, let us resolve the gravitational force. Gravitational force can be resolved into two components relative to the incline.

Parallel to the incline: mg sinθ

Perpendicular to the incline: mg cosθ

Normal force (N) equals the perpendicular component of the gravitational force in its magnitude but acts opposite in direction to it.

N= mg cosθ

The kinetic friction force is given by: f_k=μ_kN= μ_k (mg cosθ)

The net force F_net acting down the incline is the difference between the component of the gravitational force parallel to the incline and the kinetic friction force.

F_net = mg sinθ -f_k

Let's substitute f_k from the equation above.

Since f_k = μ_k (mg cosθ)

F_net = mg sinθ - μ_k (mg cosθ)

Factor out mg

F_net = mg (sinθ- μ_kcosθ)

According to Newton's 2^nd law, the net force F_net equals the mass times the acceleration a

F_net =ma

Substituting F_net from the above:

mg (sinθ- μ_k cosθ) = ma

Let's cancel m from both sides of the equation:

g(sinθ- μ_kcosθ)=a

The acceleration a is: a= g(sinθ- μ_kcosθ)

Therefore, concluding the above, the expression a= g(sinθ- μ_kcosθ) shows that the object’s acceleration down the incline is dependent on the gravitational constant g, the angle of the incline θ, and μ_k that is the kinetic friction coefficient. This expression is independent of the specific object’s mass, which confirms that the acceleration does not depend on the mass.