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 39 kg . One child pulls with a constant force of 97 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? What is the work done on the tricycle by the two ropes if the children pull constantly like this while the tricycle moves 2.0 m?

Explanation / Answer

97 N towards right at an angle of 45 degrees

Hence

The vector can be written as F1 = 97 cos 45 i + 97 sin45 j

I being the left to right axis and j being the back to front axis

Let the other child pulls with the given force at an angle of a with the left.

Hence to balance this out 80 cos a = 97 cos 45

cos a = (97 cos 45)/80

a = 30.98°

Hence total force on the tricycle taking it forward = 97 sin 45 + 80 sin 30.98° = 123.717 N

Total Inertia = 39 kg

Hence

F = ma

123.717 = 39* a

a = 3.172 m/s^2

Work Done = 123.717*2 = 247.434 J

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