Question 142: Centre of Mass and Rotational Motion

Question

A heavy iron bar of weight $$12 \mathrm{~kg}$$ is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle $$60^{\circ}$$ with the horizontal, the weight experienced by the man is :

Accepted Answer

First, let us restate the problem in our own words:
We have a uniform iron bar of weight $12\,\mathrm{kgf}$ (meaning its mass is $12\,\mathrm{kg}$, so its weight is $12g\,\mathrm{N}$ in SI units).
One end of the bar is on the ground (assume a frictionless contact), and the other end is on a manÃ¢ÂÂs shoulder (also assume frictionless contact).
The bar makes an angle of $60^\circ$ with the
horizontal
.
We want to find how much load (Ã¢ÂÂeffective weightÃ¢ÂÂ) the man feels on his shoulder.
Under these (common) assumptions of frictionless contacts at both ends,
each contact force is perpendicular to the bar
. We can solve the problem by taking torques about the ground contact point.
1. Labeling and geometry
Let the length of the bar be $L$.
The bar is inclined at $60^\circ$ from the horizontal.
Because the bar is uniform, its center of gravity (C.O.G.) lies at its midpoint, i.e., at a distance $L/2$ from either end.
Let the weight of the bar be $W = 12g$.
Let the reaction force from the manÃ¢ÂÂs shoulder be $R$. (Magnitude unknown.)
Let the reaction force at the ground be $R_g$. (We do not actually need its exact value to answer the question.)
We will place our origin (for torque calculation) at the ground contact point.
2. Torques about the ground contact
Torque due to the barÃ¢ÂÂs weight
The weight $W = 12g$ acts
downward
at the midpoint of the bar.
The position vector from the pivot (ground) to the barÃ¢ÂÂs midpoint is $	frac{L}{2}$ long and makes $60^\circ$ with the horizontal.
The weight is vertical downward, i.e., $90^\circ$ from the horizontal.
The angle between the position vector ($60^\circ$ from horizontal) and the weight ($90^\circ$ from horizontal) is
$       90^\circ - 60^\circ \;=\; 30^\circ.     $
Hence, the torque magnitude from the weight is
$       	au_W \;=\; \bigl(12g\bigr)\,\bigl(	frac{L}{2}\bigr)\,\sin(30^\circ)                  \;=\; 12g \cdot 	frac{L}{2}\cdot 	frac12                  \;=\; 12g \cdot 	frac{L}{4}.     $
Torque due to the manÃ¢ÂÂs reaction force $R$
The manÃ¢ÂÂs shoulder is at the other end of the bar, i.e., $L$ from the pivot.
For frictionless contact, $R$ is perpendicular to the bar. Since the bar is at $60^\circ$ from horizontal, a line perpendicular to the bar is $90^\circ$ away from the barÃ¢ÂÂs direction, so the angle between the barÃ¢ÂÂs position vector ($\vec{r}$) and $R$ is $90^\circ$.
Thus, the torque from $R$ about the ground contact is
$       	au_R \;=\; R \,	imes\, L \,	imes\, \sin(90^\circ)                   \;=\; R \cdot L.     $
Because the bar is in static equilibrium, the net torque about the ground contact must be zero. Taking the
clockwise
direction (by weight) as positive, we have
$   	au_W \;-\; 	au_R \;=\; 0    \quad\Longrightarrow\quad    12g \cdot \frac{L}{4} \;=\; R \cdot L. $
Solving for $R$,
$   R \;=\; \frac{12g}{4}          \;=\; 3g. $
Hence,
the force on the manÃ¢ÂÂs shoulder
has magnitude $3g$ in newtons.
3. Converting to Ã¢ÂÂkg weightÃ¢ÂÂ
Often, such problems phrase the answer in Ã¢ÂÂkg of weightÃ¢ÂÂ rather than newtons. Since $W = mg$ for a mass $m$ in EarthÃ¢ÂÂs gravity, a force of $3g\,\mathrm{N}$ corresponds to a Ã¢ÂÂweightÃ¢ÂÂ of $3\,\mathrm{kgf}$.
So the man effectively
feels 3 kg
of load on his shoulder.

Answer

$$3 \mathrm{~kg}$$

Answer

$$6 \mathrm{~kg}$$

Answer

$$6 \sqrt{3} \mathrm{~kg}$$

Answer

$$12 \mathrm{~kg}$$

Centre of Mass and Rotational Motion

A heavy iron bar of weight $12 \mathrm{~kg} is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle 60^{\circ}$ with the horizontal, the weight experienced by the man is :

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