Question

Force F acts at point O. Let F1 and F2 denote the components of F such that F = F1 + Fつ. The figure shows component F 1* F 4 lb Fi = 6 lb 80° 24° 0 Determine the magnitude in pounds and direction in degrees counterclockwise from the +x-axis of component F2 magnitude direction Irg counterclockwise from the +x-axis

Answer 1

Figure-(2) Write the expression for the net force along the horizontal axis F cos 80Rcos 24°Fcos 6 Substitute the given values in the above expression. 4 cos 800-6 cos 24° + F, cos θ F, cos θ 4 cos 800-6cos 240 F, cos θ -4.787 Write the expression for the net force along the vertical axis F sin 80+ Fsin 24°F sin 0 E sin θ =-(F sin 80° + E sin 24°) Substitute the given values in the above expression. E sin θ -(4sin 80°16 sin 24°) F sin -6.38 Divide equation (2) by equation (1) 6.38 4.787 tan e- 0.38 e tan-6.38 4.787 e 53.12° Substitute 53.120 for e in the equation (1)

F cos (5312) 4.787 -4.787 cos (53.12) The negative sign indicates that the direction of force is opposite to chosen direction. Write the direction of force F from the positive direction of the x axis. e (360 53.12) 306.88 Thus, the magnitude of the force F is 7.98 1b and direction is 306.889