Bio physics help In the following figure, the leg of a locust is composed of a f

Question

Bio physics help In the following figure, the leg of a locust is composed of a femur and the tibia. The two bones meet at the tibia joint O, where two muscles meet called the extensor and flexor muscles. The tendon lock AB holds these bones in the proper position so that a locust can jump. Let the distance OD be 0.76 mm, the distance OA be 1.64 mm, the angle ODE be 144°, and OAC be 25o. Given the magnitude of the forces IDE! = 0.100 N and IAC|-0.300 N, determine the resultant torque. If the moment of inertia is 0.0003 kg-m2, then what is the angular acceleration of the locust's leg? Femur Tibia F

Explanation / Answer

given
magnitude of forces
|DE| = 0.1 N
|AC| = 0.3 N
OD = 0.76 mm = 0.76*10^-3 m
OA = 1.64 mm = 1.64*10^-3 m
Angle ODE = 144 deg
Angle OAC = 25 deg

so net torque is given by
T = |AC|*OA*sin(OAC) - |DE|*OD*sin(ODE) [ taking counterclockwise torque to be positive]
T = 0.3(1.64*10^-3)*sin(25) - 0.1*(0.76*10^-3)*sin(144)
T = 0.16325*10^-3 Nm

now moment of inertia, I = 0.0003 kgm^2
then angular acceleration alpha = ?
alpha*I = T
alpha = 0.16325*10^-3/0.0003 = 0.544188 rad/s/s

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