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In a tug-of-war game on one campus, 15 students pull on a rope at both ends in an effort to displace the central knot to one side or the other. Two students pull with force 190 N each to the right, four students pull with force 92 N each to the left, five students pull with force 64 N each to the left, three students pull with force 158 N each to the right, and one student pulls with force 260 N to the left. Assuming the positive direction to the right, express the net pull on the knot in terms of the unit vector. How big is the net pull on the knot? In what direction?

Respuesta :

abidemiokin

Answer:

94N to the left

Explanation:

Net force will be the net pull on the knot

Net force = Sum of Direction of force towards the right - Sum of Direction of force towards the left.

For direction of force towards the right:

Two students pull with force 190 N = 2(190) = 380N

Three students pull with force 158 N each to the right = 3(158) = 474N

Total sum of force towards the right = 380N+474N = 854N

For direction of force towards the left:

four students pull with force 92 N each to the left = 4(92) = 368N

Five students pull with force 64 N each to the left = 5(64) = 320N

One student pulls with force 260 N to the left = 1(260) = 260N

Total sun for forces towards the left = 368N+320N+260N = 948N

Net force = 854N - 948N

Net force = -94N

Hence the net pull on the knot is -94N

The net pull on the knot is 94N (Its magnitude) to the left (due to the negative value we arrived at)

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