Asked by Victor
Given the vectors below, determine the resultant and the equilibrant using trigonometry. Show all your steps.
330.0 newtons @ 125 degrees
250.0 newtons @ 60.0 degrees
330.0 newtons @ 125 degrees
250.0 newtons @ 60.0 degrees
Answers
Answered by
Henry
Fr = 330[125o] + 250[60o] = Resultant force.
X = 330*Cos125 + 250*Cos60 = -64.3 N.
Y = 330*sin125 + 250*sin60 = 487 N.
Q2.
Tan A = Y/X = 487/-64.3 = -7.57387.
A = -82.5o = 82.5o N. of W. = 97.5o CCW.
Fr=Y/sin A = 487/sin97.5 = 491 N.[97.5o}
Fr + Fe = 0, Fe = -Fr.
Fe = -(-64.3 + 487i) = 64.3 - 487i =
491[82.5o] S. of E. = 491 N.[277.5o]CCW.
= Equilibrant force.
Or Fe=-Fr=Fr[97.5o+180o] = 491N.[277.5]
X = 330*Cos125 + 250*Cos60 = -64.3 N.
Y = 330*sin125 + 250*sin60 = 487 N.
Q2.
Tan A = Y/X = 487/-64.3 = -7.57387.
A = -82.5o = 82.5o N. of W. = 97.5o CCW.
Fr=Y/sin A = 487/sin97.5 = 491 N.[97.5o}
Fr + Fe = 0, Fe = -Fr.
Fe = -(-64.3 + 487i) = 64.3 - 487i =
491[82.5o] S. of E. = 491 N.[277.5o]CCW.
= Equilibrant force.
Or Fe=-Fr=Fr[97.5o+180o] = 491N.[277.5]
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