Respuesta :
Answer:
(a) Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
(b) Magnitude of resulting torque = = 3.99 Nm
(c) angular acceleration = = 0.009027 rad/s²
Explanation:
Given Data;
I = 442 kg˙m2
rx = 0.76 m,
ry = 0.035 m,
rz = 0.015 m,
Fx = 3.6 N,
Fy = -2.8 N,
Fz = 4.4 N
F = Fx i + Fy j + Fz ------------------------------1
r = rx i + ry j + rz k ------------------------------2
(a) Torgue is given by the formula;
T = r * F ------------------------------------3
Putting equation 1 and 2 into equation 3, we have;
Torque= r x F
= (rx i +ry j +rz k) x (Fx i + Fy j +Fz k )
= (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
Therefore,
Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
b)
Putting given values into the above expression, we have
torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
=(0.035*4.4 - (0.015*-2.8))i +(0.015*3.6 - 0.76*4.4)j+(0.76* -2.8 - 0.035*3.6)k
= (0.154 +0.041) i + (0.054 - 3.344) j + (-2.128 -0.126) k
= (0.196) i - (3.29) j + (-2.254) k
Magnitude of resulting torque = √(0.196² + 3.29² +2.254²
=√15.943031
= 3.99 Nm
c) Angular acceleration is given by the formula;
angular acceleration = torque/moment of inertia
= 3.99/ 442
= 0.009027 rad/s²