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Chapter 3: Problem 28

A 26-ton Navy jet (Fig. 3-27) requires an air speed of \(280 \mathrm{ft} / \mathrm{s}\) for lift-off. Its own engine develops a thrust of \(24,000 \mathrm{lb}\). The jet is to take off from an aircraft carrier with a \(300-\mathrm{ft}\) flight deck. What force must be exerted by the catapult of the carrier? Assume that the catapult and the jet's engine each exert a constant force over the 300 -ft takeoff distance.

Short Answer

Expert verified
The catapult must exert a force of approximately \(186927.57 lb\).

Step by step solution

01

Determine the Acceleration Required

We can use the kinematic equation \(v^2 = u^2 + 2as\) where \(v\) is the final speed (280 ft/s), \(u\) is the initial speed (0 ft/s), \(a\) is the acceleration, and \(s\) is the distance (300 ft). Solving for \(a\) gives \(a = (v^2 - u^2) / (2s)\). Substituting the known values, we get \(a = (280^2 - 0) / (2 * 300) = 130.67 ft/s^2.
02

Calculate the Total Force Needed

We use the formula \(F = ma\), where \(m\) is the mass of the jet and \(a\) is the acceleration determined in the previous step. The mass needs to be converted to slug (mass unit in the Imperial system) from tons for unit consistency. 1 ton = 2000 lb, and 1 slug = 32.2 lb. So, \(m = 26 * 2000 / 32.2 = 1615.53 slug\). Substituting the values in the equation gives \(F = 1615.53 * 130.67 = 210927.57 lb\). This is the total force required for the jet to lift off.
03

Find the Force Exerted by the Catapult

The total force required for lift-off is the sum of the forces exerted by the jet engine and the catapult. To find the force exerted by the catapult, we subtract the force from the jet engine from the total force required. So, the force exerted by the catapult is \(210927.57 lb - 24000 lb = 186927.57 lb\).

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Most popular questions from this chapter

In a modified tug-of-war game, two people pull in opposite directions, not on a rope, but on a \(25-\mathrm{kg}\) sled resting on an icy road. If the participants exert forces of \(90 \mathrm{~N}\) and \(92 \mathrm{~N}\), what is the acceleration of the sled?

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A car moving initially at a speed of \(50 \mathrm{mi} / \mathrm{h}(\approx 80 \mathrm{~km} / \mathrm{h})\) and weighing \(3000 \mathrm{lb}(\approx 13,000 \mathrm{~N})\) is brought to a stop in a distance of \(200 \mathrm{ft}(\approx 61 \mathrm{~m})\). Find \((a)\) the braking force and \((b)\) the time required to stop. Assuming the same braking force, find \((c)\) the distance and \((d)\) the time required to stop if the car were going \(25 \mathrm{mi} / \mathrm{h}(\approx 40 \mathrm{~km} / \mathrm{h})\) initially.
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