Answer:
- Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
- Magnitude of resulting torque = 3.84 Nm
- Angular acceleration = = 0.0086877 rad/s² rad/s²
Explanation:
It is given that:
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 (equation 1)
r = rx i + ry j + rz k (equation 2)
(a) Torque
T = r * F (equation 3)
by 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) Magnitude of the torque
Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
=(0.025 x 2.2 - (0.035 x (-1.2))) i +(0.035 x (-7.8)) - 2.19 x 2.2) j+(2.19 x 1.2 - 0.025 - (-7.8)) k
= (0.055 - 0.042) i + (0.273 - 4.818) j + (-2.628 -0.195) k
= (0.013) i - (-4.545) j + (-2.433) k
Magnitude of resulting torque = √0.013² + 4.545² - 2.433²
=√14.751
= 3.84 Nm
c) Angular acceleration
Since
angular acceleration = torque/moment of inertia
= 3.84/ 442
= 0.0086877 rad/s²
Hence,
- Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
- Magnitude of resulting torque = 3.84 Nm
- Angular acceleration = = 0.0086877 rad/s² rad/s²
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