Two children pull a third child on a tricycle by means of two ropes tied to the

Question

Two children pull a third child on a tricycle by means of two ropes tied to the handlebars. The combined inertia of the child and tricycle is 35 kg . One child pulls with a constant force of 95 N directed to the right of the straight-ahead direction at an angle of 45 . If the second child pulls with a force of 80 N , at what angle to the left of the straight-ahead direction must he pull to make the tricycle move straight ahead? With the second child pulling at the angle you calculated in part A, what is the acceleration of the tricycle?

Explanation / Answer

F1x = F1*cos45

F1y = F1*sin45

F2x = F2*costheta

F2y = -F2*sintheta

the tricycle is moving along x

Fnety = 0

F1y + F2y = 0

F1*sin45 = F2*sintheta

95*sin45 = 80*sintheta

theta = 58.02 degrees

B)

Fnet = F1x+ F2x = (95*cos45)+(80*cos58.02) = 107.77 N

acceleration = a =Fnet/m = 107.77/35 = 3.079 m/s^2

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