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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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Rewritten by : Barada
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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