F1+F2+F3=0 (equilibrium)
10E+5cos60N+5Sin60E+F3=0
solve for F3
body is at equilibrium under the action of three forces.one force is 10N acting due east and one is 5N in the direction 60¡ã northeast . what is the magnitude and direction of the third force
4 answers
F1+F2 = 10 + 5[60].
F1+F2 = 10+5*Cos60 + i5*sin60,
F1+F2 = 12.5 + 4.33i. = 13.2N[19.1o]N. of E.
F3 is equal in magnitude to F1+F2 but 180o out of phase(opposite).
F3 = 13.2N[19.1+180] = 13.2N[199.1]CCW = 13.2N[19.1o] S. of W.
F1+F2 = 10+5*Cos60 + i5*sin60,
F1+F2 = 12.5 + 4.33i. = 13.2N[19.1o]N. of E.
F3 is equal in magnitude to F1+F2 but 180o out of phase(opposite).
F3 = 13.2N[19.1+180] = 13.2N[199.1]CCW = 13.2N[19.1o] S. of W.
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