A 34,000-lb airplane lands on an aircraft carrier and is caught by an arresting cable. The cable is inextensible and is paid out at A and B from mechanisms located below deck and consisting of pistons moving in long oil-filled cylinders. The piston-cylinder system is designed to maintain a constant tension in the cable. Knowing that the landing is 110 mi/h and the airplane travels a distance d = 90 ft after being caught, determine the tension in the cable.

  F = 153637 lb

Explanation:

Let's use the kinematic equations to find the airplane's braking speed

       v² = v₀² + 2 a d

The final speed is zero

       a = - v₀² / 2d

Let's reduce the magnitudes

       v₀ = 110 mi / h (5280 ft / 1 mile) (1 h / 360 s) = 161.33 ft / s

       a = -161.33²/2 90

       a = -144.60 ft / s²

Now we can use Newton's second law to find the tension

       F = m a

       F = (34000/32) 144.60

      F = 153637 lb

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y