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saw5 [17]
2 years ago
6

Three people pull simultaneously on a stubborn donkey. jack pulls eastward with a force of 92.5 n, jill pulls with 89.9 n in the

northeast direction, and jane pulls to the southeast with 163 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.
Physics
1 answer:
alekssr [168]2 years ago
7 0
Jack------------ force of 92.5 n   eastward-------Fjack(X)=92.5 n   Fjack(Y)=0

<span>jill ------------------------------- force of 89.9 n   northeast
Fjill(X)=cos45*89.9=63.57 n
</span>Fjill(Y)=sin45*89.9=63.57 n<span>

</span>jane -----------------------------force of 163 n   southeast
Fjane(X)=cos45*163=115.26 n
Fjane(X)=-sin45*163=-115.26 n

Ftotal (X)=92.5+63.57+115.26=271.33 n
Ftotal (Y)=0+63.57-115.26=-51.69 n

Fotal=((271.33)^2+(-115.26)^2) ^0.5=294.80 n southeast

the magnitude of the net force the people exert on the donkey. is 294.80 n southeast
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