Solution: This solution uses an origin located at the center of the lower can. The right triangle has a hypotenuse of 2r. The horizontal leg has a length of (w – 2r). Using Pythagorean Theorem, the vertical leg is: dvert = √√(2r)² - (w – 2r)² – - Square out the 2nd term and clean up: dvert = √√4wr w² - Torque balance equation: TCCW - TCW = 0 Let 2(mg) be the total weight of both cans, which can be viewed as acting at the CM of the 2 cans. Plug in using Tr₁F: w-2r = 0 (Nwall) (√4wr – w²) – (2(mg)) (1-2) = - Cans in a Box N wall W CMQ 2(mg) N wall N floor Jamie puts 2 beverage cans in a cooler. Ignore friction. Each can has a mass m and a radius r. The width of the cooler is w. Each wall exerts a normal force Nwall as shown. Hint: Use the right triangle shown. Choose an origin and calculate torques. 9 10 r พ m Nwall 0.491 3.12cm 10.5cm N kg Type your answer... r พ m Nwall 3.42cm 10.1cm 4.13N kg 0.777