Question

Find the tension in the string connecting block of mass $M$ and ring of mass $m$ at the instant shown. Assume all surfaces to be frictionless and the ring is constrained to move along the wire only. $M=25\text{kg};m=9\text{kg}$ and $g=10{\text{m/s}}^2$

• A. $50\text{N}$
• B. $100\text{N}$
• C. $125\text{N}$
• D. $150\text{N}$

Answer 1

The correct option is C $125\text{N}$

From constrained condition, using method of intercept we have
${\stackrel{..}{L}}_1+{\stackrel{..}{L}}_2=0$
$\Rightarrow -a_1\cos \theta +a_2=0$
$\Rightarrow a_2=a_1\cos \theta \dots \dots \left(1\right)$
From FBDs of the ring and mass

$T\cos \theta =m{a}_1\dots \dots \left(2\right)$
$Mg-T=M{a}_2\dots \dots \left(3\right)$
From (1), (2) and (3)
$(g-\frac{T}{M})=\frac{T{\cos }^2\theta }{m}\Rightarrow T[\frac{{\cos }^2\theta }{m}+\frac{1}{M}]=g$
$\Rightarrow T\left(\frac{M{\cos }^2\theta +m}{Mm}\right)=g\Rightarrow T=\frac{Mmg}{m+M{\cos }^2{53}^{\circ }}$
$=\frac{25\times 9\times 10}{9+25\times \frac{9}{25}}=125\text{N}$