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Q 7.22    As shown in Fig.7.40, the two sides of a step ladder $BA$ and $CA$ are $1.6m$ long and
   hinged at A. A rope $DE$ , $0.5m$ is tied half way up. A weight $40 kg$ is suspended from
   a point $F$ , $1.2m$ from $B$ along the ladder $BA$ . Assuming the floor to be frictionless
   and neglecting the weight of the ladder, find the tension in the rope and forces
   exerted by the floor on the ladder. $(Take g=9.8m/s^{2})$ .
(Hint: Consider the equilibrium of each side of the ladder separately.)

Answers (1)

best_answer

The FBD of the figure is shown below :

Rotational motion, 20149

Consider triangle ADH,

$AH\ =\ \sqrt{\left ( AD^2\ -\ DH^2 \right )}$ (AD and DH can be found using geometrical analysis.)

$=\ \sqrt{\left ( 0.8^2\ -\ 0.25^2 \right )}\ =\ 0.76\ m$

Now we will use the equilibrium conditions :

(i) For translational equilibrium :

$N_c\ +\ N_b\ =\ mg\ =\ 40\times 9.8\ =\ 392\ N$ ......................................(i)

(ii) For rotational equilibrium :

$-N_b\times BI\ +\ mg\times FG\ +\ N_c \times CI \ +\ T\times AG\ -\ T\times AG\ =\ 0$

or $(N_c\ -\ N_b)\times 0.5\ =\ 49$

or $N_c\ -\ N_b\ =\ 98$ ............................................................(ii)

Using (i) and (ii) we get :

$N_c\ =\ 245\ N$ and $N_b\ =\ 147\ N$

Now calculate moment about point A :

$-N_b \times BI \ +\ mg\times FG\ +\ T\times AG\ =\ 0$

Solve the equation : $T\ =\ 96.7\ N$

Posted by

Devendra Khairwa

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