Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 77.3 N, Jill pulls with 61.7 N in a direction 45° to the left, and Jane pulls in a direction 45° to the right with 147 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. Magnitude of the net force: _N What is the direction of the net force? Express this as the angle from straight ahead between 0° and 90°, with a positive sign for angles to the left and a negative sign for angles to the right.
Direction of the net force: _°

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aristocles aristocles · Answer 1 · 2019-10-11T03:22:50+02:00

Answer:

[tex]F_{net} = 232.8 N[/tex]

towards right so it is -15 degree

Explanation:

Net force in forward direction due to all three is given as

[tex]F_x = F_1 + F_2cos45 + F_3cos45[/tex]

here we know that

[tex]F_1 = 77.3 N[/tex]

[tex]F_2 = 61.7 N[/tex]

[tex]F_3 = 147 N[/tex]

[tex]F_x = 77.3 + 61.7 cos45 + 147 cos45[/tex]

[tex]F_x = 224.9 N[/tex]

Similarly in Y direction we will have

[tex]F_y = F_3 sin45 - F_2 sin45[/tex]

[tex]F_y = (147 - 61.7)sin45[/tex]

[tex]F_y = 60.3 N[/tex]

Now the net force on the donkey is given as

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

[tex]F_{net} = \sqrt{224.9^2 + 60.3^2}[/tex]

[tex]F_{net} = 232.8 N[/tex]

Now direction of force is given as

[tex]tan\theta = \frac{F_y}{F_x}[/tex]

[tex]tan\theta = \frac{60.3}{224.9}[/tex]

[tex]\theta = 15^o[/tex] towards right so it is -15 degree