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A non-uniform rod $AB$ has mass $6\mathrm{kg}$ and length $6L$. Its centre of mass is at a distance $L$ from $A$. One end of a light inextensible string is attached to the rod at A and the other end is attached to a small smooth ring $R$ of mass $2\mathrm{kg}$ threaded on the rod. The ring is also attached to B by a light elastic spring of modulus $\lambda$ and natural length $\frac{L}2$.

​ The rod hangs in equilibrium with the inextensible string passing over a smooth peg $P$ and with the ring a distance $L$ from $B$. The portion $AP$ of the inextensible string is perpendicular to the rod.

Find the value of $\lambda$ for this system.

Below is my attempt at a solution with a diagram. I am wondering whether my issue is with Normal reaction force $2$.

Please could I get a hint, and not a full solution.

[![Here is an annotated diagram] enter image description here

Force of spring: $F_s=\frac{\lambda}{L_0}x$ Parallel to the rod: $T\sin⁡\alpha-8\mathrm{g}\sin\theta=F_s=\frac{\lambda}{\left(\frac{L}2\right)}\times\frac{L}2\Rightarrow T\sin⁡\alpha-8\mathrm{g}\sin\theta=\lambda$

$N_1=T\cos\alpha$ and $N_2=T$

Vertically: $T\cos\theta+T\cos⁡(\alpha-\theta)=8\mathrm{g}+N_1 \cos⁡\theta+N_2\cos⁡\theta =8\mathrm{g}+T\cos\alpha \cos\theta+T\cos\theta\\\Rightarrow T\cos\theta+T(\cos⁡\alpha\cos⁡\theta+\sin⁡\alpha\sin⁡\theta)=8\mathrm{g}+T\cos\alpha\cos\theta+T\cos\theta\\\Rightarrow T\sin\alpha\sin⁡\theta=8\mathrm{g}$

Moments about $A$: $L(6\mathrm{g}\cos\theta)+5L(2\mathrm{g}\cos\theta+N_1)=5L(T\cos⁡\alpha)\\\Rightarrow(6\mathrm{g}\cos\theta)+5(2\mathrm{g}\cos\theta)=0\\\Rightarrow\theta=90^{\circ}\\\Rightarrow T\sin\alpha=8\mathrm{g}$

Horizontally: $T\sin\theta=N_1\sin⁡\theta+N_2\sin⁡\theta+T\sin(\alpha-\theta)\\\Rightarrow T=T\cos\alpha+T+T\sin(\alpha-90^{\circ})\\\Rightarrow T\cos\alpha=T\cos\alpha$

This is not very useful.

JCP13321
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1 Answers1

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Since you only want hints and not a complete solution:

The key to the problem is to understand what forces apply to the system as a whole and what forces are internal and which will therefore apply to either the ring on its own or to the rod on its own.

For the system as a whole, the only forces you should be showing are the tensions (which must have equal magnitude since $P$ is smooth) and the two weights $6g$ and $2g$.

By taking moments at $A$ and resolving forces in the directions along the rod and perpendicular to it, you can establish that $$\cos \alpha=\frac23$$ and that $$\tan\theta =\frac{1}{\sqrt{5}}.$$ You can also calculate the tension, but that isn’t really necessary as it turns out.

For the ring alone you should show the tension, the normal reaction perpendicular to the rod, the weight of the ring, and the spring force (which you seem to have ignored). You can use Hooke’s Law to establish that this force has magnitude $\lambda$, the quantity you are trying to find.

Resolve forces parallel to the rod, to avoid having to involve the normal reaction, and you will get the answer fairly quickly.

I got $$\lambda=g\sqrt{6}$$

Edit

As requested by the OP, here are the steps:

Taking moments for the whole system about $A$: $$T\times 5L\cos\alpha=6g\cos\theta+2g\times 5L\cos\theta$$ $$\implies T\cos\alpha=\frac{16}{5}g\cos\theta. (1)$$

Resolving forces for the whole system along $AB$: $$T\sin\alpha=8g\sin\theta. (2)$$

And in the direction $AP$, $$T+T\cos\alpha=8g\cos\theta. (3)$$

Equations $(1)$ and $(3)$ give $$T=8g\cos\theta-\frac{16}{5}g\cos\theta = \frac{24}{5}g\cos\theta$$

So now equation $(1)\implies \cos\alpha=\frac23$

Then equation $(2)$ divided by equation $(3)$ gives $$\tan\theta=\frac{\sin\alpha}{1+\cos\alpha}=\frac{1}{\sqrt{5}}$$

Finally, resolving forces for the ring alone in the direction $AB$ gives $$T\sin\alpha=\lambda+2g\sin\theta$$

So from equation $(2)$, $$\lambda=6g\sin\theta$$ and $\sin\theta=\frac{1}{\sqrt{6}}$, so we get $$\lambda=g\sqrt{6}$$

David Quinn
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