Part A What is the correct expression of the internal torque in segment AB? A shaft is fixed at A. A clockwise distributed torqu
e with magnitude 2 kN-m/m is applied between A and B. A counterclockwise torque of 400 N-m is applied at B, and a clockwise torque of 600 N-m is applied at the free-end C. The segments AB and BC have lengths 0.8 m and 0.6 m, respectively. What is the correct expression of the internal torque in segment AB? A shaft is fixed at A. A clockwise distributed torque with magnitude 2 kN-m/m is applied between A and B. A counterclockwise torque of 400 N-m is applied at B, and a clockwise torque of 600 N-m is applied at the free-end C. The segments AB and BC have lengths 0.8 m and 0.6 m, respectively. 2000x + 1400 N-m 2000x N-m –2000x +1800 N-m –2000x – 1400 N-m 2000x – 1800 N-m –2000x – 200 N-m 2000x + 200 N-m –2000x N-m
The torque at a given point is determined as the perpendicular force multiply by the distance of the force from the given point. The conventional method is to take clockwise direction as positive and anticlockwise direction as negative. Therefore:
Torque at the segment AB = 2000*x + 400 = (2000x + 400) N-m
In conclusion, projectiles travel with a parabolic trajectory due to the fact that the downward force of gravity accelerates them downward from their otherwise straight-line, gravity-free trajectory.
