High School

Three people pull simultaneously on a stubborn donkey.

- Jack pulls directly ahead of the donkey with a force of 98.3 N.
- Jill pulls with 62.5 N in a direction 45° to the left.
- Jane pulls in a direction 45° to the right with 117 N.

Calculate the net force exerted on the donkey and determine its direction.

Answer :

The magnitude of the net force the people exert on the donkey is 239 N.

Net force :

[tex]F_{n}[/tex]= [tex]F_{1}+F_{2} +F_{3}[/tex]

Now finding the x and y components of the forces F2 and F3,

[tex]F_{x}[/tex] = F1 + F2cos45°+F3sin45°

[tex]F_{x}[/tex] = 98.5 + 69.3 ×√2/2 + 125 ×√2/2

[tex]F_{x}=235.9N[/tex]

[tex]F_{y}[/tex] = -F2sin45°+F3sin45°

= −69.3 ×√2/2+ 125 ×√2/2

[tex]F_{y}[/tex] = 39.4 N

Now find the magnitude

[tex]F_{n}[/tex] = √235.9^2+39.4^2 = 239 N

The magnitude of the net force the people exert on the donkey is 239 N.

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Jane and Jill are stubborn because they have opposite force directions.

We need to know about force resultant to solve this problem. The force resultant is the total net force applied to the object according to the direction. It can be written as

R = F1 + F2 + ... + Fn

where R is force resultant (net force)

From the question above, we know that

F Jack = 98.3 N

F Jill = 62.5 N

F Jane = 117 N

Assume that the direction

Jack = east

Jill = north east

Jane = south east

The Force to the north direction is

F Jill y = Fjill . cos45

F Jill y = 62.5 . √2/2

F Jane y = -F Jane . cos45

F Jane y = -117 . √2/2

Jane and Jill are stubborn because they have opposite directions.

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