Bob (80 kg, treat as point mass) is trying to figure out the strength of a rope,
ID: 1624029 • Letter: B
Question
Bob (80 kg, treat as point mass) is trying to figure out the strength of a rope, using a seesaw (3 m long, approximately massless, with hinge at its center). He ties a piece of the rope from the ground to one end of the seesaw, and then finds that if he sits slightly over 0.5 m from the pivot (on the other side), the rope breaks. In the above diagram, please draw in all external forces that act on the Bob seesaw system, Make sure to indicate directions, and label what each one is. What is the tension at which the rope breaks? What will be the angular acceleration of the system at the moment the rope breaks? What minimum force must the triangular support of the seesaw be able to withstand, if the tope is to break first? Suppose now that the seesaw plank is 30 kg (not massless). Repeat part c, and then compare your new result to previous (conceptually explain if/what difference you find).Explanation / Answer
b) when the rope breaks, tension in the rope = T
From moment balance about the centER of see saw
T*1.5 = 80*9.81*0.5
T = 261.6 N
c) When rope breaks, net moment = T*1.5 Nm = I*alpha
I = 80*0.5^2
so, alpha = 19.62 rad/s/s
d) This force = T + mg = 261.6 + 80*9.81 = 1046.4 N
e) T*1.5 Nm = I*alpha
I = 80*0.5^2 + 30*3^2/12
alpha = 9.232 rad/s/s