Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 61.3 N, Jill pulls with 83.9 N in a direction 45° to the left, and Jane pulls in a direction 45° to the right with 137 N. (Since the donkey is involved with such uncoordinated people, who can blame it for being stubborn?) Find the magnitude of the net force the people exert on the donkey.

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aristocles aristocles · Answer 1 · 2019-07-29T02:42:37+02:00

Answer:

[tex]F_{net} = 220.8 N[/tex]

Explanation:

It is pulled by three forces as given below

1. Jack pulls directly ahead of the donkey with a force of 61.3 N,

2. Jill pulls with 83.9 N in a direction 45° to the left, and

3. Jane pulls in a direction 45° to the right with 137 N.

Now net force directly in forward direction given as

[tex]F_x = 61.3 N + 83.9 cos45 + 137cos45[/tex]

[tex]F_x = 217.5 N[/tex]

Now similarly in perpendicular to this we have

[tex]F_y = 137 sin45 - 83 sin45 [/tex]

[tex]F_y = 38.2 N[/tex]

Now net force is given by them

[tex]F_{net} = \sqrt{F_x^2 + F_y^2}[/tex]

[tex]F_{net} = \sqrt{217.5^2 + 38.2^2}[/tex]

[tex]F_{net} = 220.8 N[/tex]