Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A small mouse has gotten into a peculiar situation. Somehow, the mouse has found

ID: 2100608 • Letter: A

Question

A small mouse has gotten into a peculiar situation. Somehow, the mouse has found himself hanging off the end of the metal structure shown in the figure below. Hopefully the structure can support the 26.6 g mouse because it would not make the fall. The structure is rigidly fixed to a wall and has mass per unit length of 5.73 g/cm and dimensions x = 21.0 cm and y = 44.1 cm, defined as y = 0.1x^2 (where the attachment point P has coordinates xP = 0, yP = 0). Calculate the torque about the attachment point P.


Explanation / Answer

total torque will be given by the torque due to mass of mouse(T1) and torque due to mass of rod itself which will act @ centroid of the given parabola(T2) T1=weight of mouse *perpendicular distance of line of action of ts weight from p(x) =26.6*10^-3*9.81*21*10^-2 =0.055 N.m arc length of parabola=integral(sqrt(1+(dy/dx)^2))dx with lower limit=0 and upper limit=21 so arc length=36.2 cm so total weight=36.2*5.73=207.4 gm x co ordinate of centroid=3*21/8=7.875 cm so T2=207.4*10^-3*9.81*7.875*10^-2=0.16 N.m T=0.16+0.055=0.215 N.m

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote