Mechanics question...

Question

Particles P and Q are attached to opposite ends of a light in-extensible string. P is at rest on a rough horizontal table. The string passes over a small smooth pulley which is fixed at the edge of the table.Q hangs vertically below the pulley (see diagram). The force exerted on the string by the pulley has magnitude $4\sqrt{2}\,\rm N$. The coefficient of friction between P and the table is $0.8$.

(i) Show that the tension in the string is $4\,\rm N$ and state the mass of Q.

Answer 1

Horizontal force on $P$ due to string balancing is the force of static friction given by $$f=\mu\cdot F_n,$$ here $F_n=mg$ where $g$ is acceleration due to gravity and $m$ is the mass of $P$.
Let the tension be $T$ in the string. Now $|T|=|f_P|=|f_Q|$ where $f_P,f_Q$ are forces on $P$ and $Q$ by the string.
Now the force on by the pulley upon the string = Tension in the vertical part of the string+Tension in the horizontal part
So $4\sqrt2= \sqrt{T^2+T^2}=\sqrt2 T\implies T=4N$ since the angle between the vertical and horizontal portion is $90^0$. The mass of $Q$ is $\dfrac{4}{g}=\dfrac{4}{9.8} kg$. Similarly mass of $P$ can be found.

Answer 2

Try to convince yourself that the free body diagram below is correct.

• $T \to$tension of the string
• $R_1 \to$Force exerted by the pulley on the string ($4\sqrt2N$ given)
• $R \to$Normal reaction force on $P$ by the table
• $\mu R \to$Static friction working on $P$

Here $\theta=45^{\circ}$

So for the following 3 systems we've got 4 equilibrium:

\begin{align} &\text{Pulley:}&R_1&=2T\cos\theta\tag{i}\\ &\text{Ball Q:}&T&=m_qg\tag{ii}\\ &\text{Ball P:}&T&=\mu R\tag{iii}\\ &\text{Ball P:}&R&=m_pg\tag{iv}\\ \end{align}

From $(i)$ we get : $$4\sqrt2=2T\cos(45^{\circ}) = \frac{2T}{\sqrt2}\\ \therefore T=4N$$ From $(ii)$ we get : $$T=m_qg=10m_q(\mathrm{taking}\ \ g=10m/s^2)\\ \therefore m_q=\frac{T}{10}=\frac{4}{10}=0.4kg$$