A small railroad engine is able to exert a 17,500 lb pulling force. The coefficient of friction for steel wheels on steel rail is 0.40. How many rail cars weighing 45,000 lb each can this engine pull on level track?
force needed per car = 0.40 (Typo ??) * 45,000 (Typo ? ) = 18,000
By the way most wheels can roll.
To find the number of rail cars the engine can pull on level track, we need to compare the pulling force of the engine to the total resistance force exerted by the rail cars.
Let's break down the problem step by step:
Step 1: Determine the resistance force exerted by each rail car:
The resistance force is the force that opposes motion and is equal to the weight of the rail car multiplied by the coefficient of friction.
Given:
Weight of each rail car (w) = 45,000 lb
Coefficient of friction (μ) = 0.40
Resistance force (R) = w * μ
R = 45,000 lb * 0.40 = 18,000 lb
Step 2: Calculate the total resistance force exerted by all the rail cars:
Since the engine is pulling multiple rail cars, the total resistance force will be the sum of the resistance forces of each rail car.
Total resistance force (R_total) = R * number of rail cars
Step 3: Compare the total resistance force to the pulling force of the engine:
We know that the pulling force of the engine is 17,500 lb.
If the pulling force of the engine is greater than or equal to the total resistance force, the engine can pull that number of rail cars. Otherwise, it will not have enough force.
Therefore, we need to find the number of rail cars that satisfies the inequality:
17,500 lb ≥ R_total
Step 4: Calculate the number of rail cars the engine can pull:
R_total = R * number of rail cars
17,500 lb = 18,000 lb * number of rail cars
Let's solve this equation for the number of rail cars:
number of rail cars = 17,500 lb / 18,000 lb
number of rail cars ≈ 0.972 rail cars
Since we can't have a fraction of a rail car, the engine can pull a maximum of 0 rail cars (or 0 whole rail cars).
Therefore, the engine cannot pull any rail cars on level track with the given conditions.