A Ferris wheel rotates three times each minute. It carries each car around a circle with the diameter of 19.0m. What force, magnitude (in N) and direction (in degrees), does the seat exert on a 35.0-kg child when the rider is halfway between the top and the bottom?

Answers

Answer 1

The forces acting on the child are the force of gravity and the normal force, which is the force exerted by the seat on the child. the force exerted by the seat on the child has a magnitude is Fn = 342.5N. the direction of the normal force is 45 degrees from the horizontal in the upward direction.

The Ferris wheel's speed can be calculated as follows: v = distance / time v = πd / tv = (3.14 * 19.0 m) / 1 min v = 60.66 m/min

The acceleration of the Ferris wheel can be calculated as follows: a = (v²) / ra = (60.66 m/min)² / (19.0 m / 2)a = 194.6 m/min²

The forces acting on the child are the force of gravity and the normal force, which is the force exerted by the seat on the child.

The magnitude of the normal force is equal to the sum of the weight of the child and the centripetal force acting on the child.

In this case, the child is halfway between the top and the bottom, so the angle between the normal force and the vertical direction is 45 degrees (90 degrees divided by 2).

As a result, the force exerted by the seat on the child has a magnitude of:

Fn = Fg + Fcf = maFn = (35.0 kg) * (9.8 m/s²) + (35.0 kg) * (194.6 m/min²)Fn = 342.5 N

The direction of the normal force is perpendicular to the seat, which is horizontal.

Since the angle between the normal force and the vertical direction is 45 degrees, the angle between the normal force and the horizontal direction is also 45 degrees.

Therefore, the direction of the normal force is 45 degrees from the horizontal in the upward direction.

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Related Questions

High-power experimental engines are being developed by the Stevens Motor Company for use in its new sports coupe. The engineers have calculated the maximum horsepower for the engine to be 630HP
. Twenty five engines are randomly selected for horsepower testing. The sample has an average maximum HP of 650
with a standard deviation of 60HP
. Assume the population is normally distributed.
Step 1 of 2 : Calculate a confidence interval for the average maximum HP for the experimental engine. Use a significance level of α=0.01
. Round your answers to two decimal places.

Answers

The 99% confidence interval for the average maximum HP for the experimental engine is (610.12, 689.88).

To calculate the confidence interval for the experimental engines' average maximum HP, we can use the following formula:

To find the z-score for α=0.01, we can refer to a standard normal distribution table or use a calculator. The z-score is approximately 2.58.

Substituting the given values into the formula, we get:

CI = 650 ± 2.58*(60/√25) CI = 650 ± 30.96

Rounding to two decimal places, the confidence interval for the experimental engines' average maximum HP is:

CI = [619.04 HP, 680.96 HP]

Therefore, we can say with 99% confidence that the true average maximum HP for the experimental engines falls between 619.04 HP and 680.96 HP. Thus, we can conclude that the experimental engines' average maximum HP is likely to be within this range. However, note that this range does not include the manufacturer's claimed maximum HP of 630 HP, which may indicate that the engines are performing below expectations.

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