Balance Problem

Relevant wiki: Torque - Equilibrium

Split the 2-kg bar in two parts. The part left of the fulcrum has a mass of 1.5 kg and a center of mass 1.5 m to the left of the pivot. The part on the right has a mass of 0.5 kg and with center of mass 0.5 m to the right of the fulcrum. Let the unknown mass of the left be $m$ . Balancing the moments we have:

$\begin{aligned} (3 \times m) + (1.5 \times 1.5) & = (0.5 \times 0.5) + (3.5 \times 1) \\ 3m + 2.25 & = 0.25 + 3.5 \\ 3m & = 1.5 \\ \implies m & = \boxed{0.5} \text{ kg} \end{aligned}$

Remember the lever effect ... and knowing that the wood mass effected on 1 meter from the base point.

$3 * m+ 1 * 2 = 1 * 3.5$

$3 * m = 1.5$

$m=0.5 Kg$

$3m+ 1 \times 2 = 1 \times 3.5$

$\Rightarrow 3m+2 = 3.5$

$\Rightarrow 3m = 3.5 - 2$

$\Rightarrow 3m = 1.5$

$\Rightarrow m = \boxed{0.5 ~ \text{kg}}$

So that the system is in equilibrium, summation of moments at $A$ must equal to zero. We have

$\sum M_A=0$ $(\text{counter-clockwise moments=clockwise moments})$

$x(3)=1.5(1.5)=3.5(1)+0.5(0.5)$

$3x+2.25=3.5+0.25$

$\color{#D61F06}\boxed{x=0.5~\text{kg}}$ $\color{#3D99F6}\boxed{\text{answer}}$

Distribution of the self-weigth (2-kg) of the beam:

Note that the $2~kg$ self-weight of the beam is acting throughout the length of the $4~m$ beam. First we divide the beam into two, $3~m$ long and $1~m$ long. Since the $2~kg$ is a uniformly distributed load, we must convert it to an equivalent point load (or concentrated load). Observed that the ratio of the lengths is $3:1$ , so the total is $1+3=4$ . That means that $\dfrac{1}{4}(2)=0.5~kg$ acts on the right side of the pin support and $\dfrac{3}{4}(2)=1.5~kg$ acts of the left side of the pin support. The $0.5~kg$ concentrated load must act at the center of the $1~m$ length beam or $0.5~m$ away from the pin support. The $1.5~kg$ concentrated load must act at the center of the $3~m$ length beam or $1.5~m$ away from the pin support.

As above figure clearly depicts that the system will have moment about point of pivot(fulcrum). Let the point be denoted as $C$ . For the system to be in equilibrium the moment to the left $\text{m}_{L}$ must be equal to the moment to right $\text{m}_{R}$ about point $C$ .

If we take a look to left portion net moment produced towards it is the sum of moment produced from left extreme and center point of beam which will lies on the mid point of beam about point $C$ . Let the required mass on the left extreme be $k$ . Then, $\begin{aligned} & \text{m}_{L} = \text{m}_{R} \\& k\times l_1+ 2\times l_2 = 3\times l_3 \\& 3k+1\times 2 = 3.5\times 1 \\& k = \boxed{0.5}\end{aligned}$

You can ignore the right 1m of wood since it is evenly split. The remaining 1m has a mass of 1kg centered 2m to the left:

3m * ? + 2m * 1kg = 1m * 3.5kg

Or, take the whole mass of the wood centered 1m to the left and get:

3m * ? + 1m * 2kg = 1m * 3.5kg

Either way,

? = 0.5kg

the torque due to the masses should be balanced.

the weight of the 2kg plank should also be considered.

torque due to plank= $\int$ x*(2/4)dx=x^2/4

let the unknown mass be x.

(x 3+3^2/4)=(3.5 1+1^2/4)

9/4+x*3=1/4+3.5

x=0.5kg

The center of mass for the 2 kg isn't really given, but you can pretty much eye ball it at 1 meter.

Torque = Force * Radius

The system is in equilibrium, so the torque on the left must equal the torque on the right.

Torque on the right:

3.5kg * 9.8m/s^2 * 1m = 34.3Nm

Torque on the left:

2kg * 9.8m/s^2 * 1m = 19.6Nm

(Note: we can represent the mass of the beam as a point in its center)

34.3Nm-19.6Nm=14.7Nm

So we need 14.7Nm of torque on the left to bring the system into equilibrium.

14.7Nm = F*(3m)

14.7Nm/3m = 4.9N

To solve for the mass use F=ma which can be rewritten as m=F/a so:

4.9N(9.8m/s^2) = m

m = 0.5kg

$\LARGE {x\times3+2\times1=3.5\times1\\ \Rightarrow3x+2=3.5\\ \Rightarrow x=\boxed{0.5}}$