A uniform thin rod of mass m = 5 kg and length L = 1.1 m can rotate about an axl

Question

A uniform thin rod of mass m = 5 kg and length L = 1.1 m can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F1 = 7 N, F2 = 5 N, F3 = 12.5 N and F4 = 15.5 N. F2acts a distance d = 0.1 m from the center of mass.

Randomized Variables

m = 5 kg
L = 1.1 m
F1 = 7 N
F2 = 5 N
F3 = 12.5 N
F4 = 15.5 N
d = 0.1 m

20% Part (a)  Calculate the magnitude ?1 of the torque due to force F1 in N?m.

20% Part (b)  Calculate the magnitude ?2 of the torque due to force F2 in N?m.

20% Part (c)  Calculate the magnitude ?3 of the torque due to force F3 in N?m.

20% Part (d)  Calculate the magnitude ?4 of the torque due to force F4 in N?m.

Explanation / Answer

apply the formula for torque acting at a point as T = Fr sin theta

where F is force and r is the distnace and theta is the angle between r and F

so

Part A : T1 = 7 * 1.1/2 = + 3.85 Nm

part B: T2 = 5 * (1.1.2-0.1)* sin 45   = + 1.59 Nm

part C: T3 = 12.5 * 0 = 0

Part D : T4 = 15.5 * 1.1/2 * sin 0 = 0

net torque T = sum of all torques

Tnet = 3.85 + 1.59 + 0 + 0

Tnet = 5.44 Nm

------------------------------

use the relation between and ang accleration A as T = I A

where I is Moment of inertia   = ML^2/12

I = 5 * 1.1 *1.1/12

I = 0.504 Kgm^2

so

Ang Accelration A = T/I

A = 5.44/0.504

A = 10.79 rad/s^2

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