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

A uniform plank of length 2.00 m and mass 25.5 kg is supported by three ropes, a

ID: 2265014 • Letter: A

Question

A uniform plank of length 2.00 m and mass 25.5 kg is supported by three ropes, as indicated by the blue vectors in the figure below. Find the tension in each rope when a685-N person is d = 0.575 m from the left end.




A 11.0-kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 11.9 m/s, determine the following.


(b) the rotational kinetic energy about its center of gravity
J

(c) its total kinetic energy
J

|1| =   N |2| = N |3| = N A uniform plank of length 2.00 m and mass 25.5 kg is supported by three ropes, as indicated by the blue vectors in the figure below. Find the tension in each rope when a685-N person is d = 0.575 m from the left end.

Explanation / Answer

when the system is in equilibrium, net force and net toruqe are zero.

Fnety = 0

T1 + T2*sin(40) - 25*9.8 - 685 = 0


T1 + T2*0.6428 - 245 - 685 = 0

T1 + 0.6428*T2 = 930 --(1)


net torque is zero


T1*cos(40)*2 - 685*0.575 - 25*9.8*1 = 0

T1 = (685*0.575 + 25*9.8*1)/(2*sin(40))

T1 = 497 N

from euqation(1)

T2 = (930-497)/0.6428

T2 = 673.7 N

Fnetx = 0

T3 = T1*cos(40)

= 497*cos(40)

= 380.72 N

2)

b) KE (rotational) = 0.5*I*w^2

= 0.5*0.5*m*r^2*w^2

= 0.25*m*v^2

= 0.25*11*11.9^2

= 389.42 J

c)

KE(total) = o.5*m*v^2 + 0.5*I*w^2

= 0.5*m*v^2 + 0.25*m*v62

= 0.75*m*v^2

= 1168.28 J

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