Question

A Plethora of Torques Due in 18 hours, 10 minutes The fiqure below is a lonq stick with evenly spaced marks. The distance between two adjacent marks is 6.30 cm к н D C G J L C For each of the following forces or set of forces determine the requested torque. Use Point K as the axis of rotation. Use the sign convention that a counterclockwise (CCW) torque is considered positive and a clockwise (CW) torque is considered negative. A force of 17.0 N is directed upward perpendicular to the stick at Point B. What is the torque from this force? 13.5N'm Submit Answer Incorrect. Tries 2/10 Previous Tries Point E. What is the torque from this force? A force of 34.0 N is directed downward perpendicular to the stick Tries 0/10 Submit Answer A force of 32.0 N is directed upward perpendicular to the stick at Point C. What is the torque from this force? Tries 0/10 Submit Answer A force of 25.0 N is directed upward at angle 35.0° to the right of the perpendicular to the stick at Point A. What is the torque from this force? Submit Answver Tries 0/10 A force of 34.0 N is applied horizontally to the right at A. What is the torque from this force? Tries 0/10 Submit Answer Two forces are applied simultaneously to the stick. A force of 17.0 N is directed upward perpendicular to the stick at Point B and a force of 32.0 N is directed upward perpendicular to the stick at Point C. What is the net torque from these two forces? Subrnit Answer Tries 0/10

Answer 1

0 ņ = of sind Y. 2x 3 cm = 12.6 (9 Fa 17N, 090° - 0:12 6 m. Y 0.126 x 17 xsin 90% = 2.142 Nm cow A 2.142 Nm YA TFsino 0.189 m, 0-90, F = 34 18:9 cun - Y = 3 X 6.3 : 1 cw T=-6.426 Nam 3 r: 3x63 cm =18.910 =0.189m; 0=48F232N TE-Tesino =-0.189 x 32x51(90) -6.043 N-m cw 0.252 m YUX6.3 cm F = 25W, O Yarfsing = 25% 0.252 x sin(90-35) ♡ = 5.161w-m cew ☺ F=34~, o=00, T=0.252m Parfsino = 0.252X34Xsin Coo) No Nam © ? = Yit P2 T= 0.126x17x singu = 2-142 Nam Tz = -0.189X32X5190'=-6.048wm wet toque q = TtYL= -3906 Norm cu Y.-3.906 W-m