As shown in the figure, the two sides of a ...

Systems of Particles and Rotational Motion

Question : 22 of 33

Marks: +1, -0

As shown in the figure, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F. 1.2 m 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 = 9.8 m s

^{-2})

(Hint : Consider the equilibrium of each side of the ladder separately.)

Solution:

Here,

BA = CA = 1.6\,\text{m}; DE = 0.5\,\text{m};

M= 40\,\text{kg}; BF = 1.2\,\text{m}

Let T = tension in the rope,

N_1, N_2 =

normal reaction at B and C respectively, i.e., forces exerted by the floor on the ladder.

In figure, we find

DE = 0.5\,\text{m}; BC = 1.0\,\text{m};

F H = \frac{1}{2} D G

= \frac{1}{2} \times 0.25 = 0.125\,\text{m}

A G = \sqrt{A D^{2} - D G^{2}}

= \sqrt{0.8^{2} - 0.25^{2}} = 0.76\,\text{m}

For translational equilibrium of the step ladder,

N_1 + N_2 - Mg = 0

N_1 + N_2 = Mg = 40 \times 9.8 = 392

...(i)

For rotational equilibrium of the step ladder, taking moments about A, we find

-N_{1} \times B K + M g \times F H + N_{2} \times C K + T \times A G - T \times A G = 0

-N_{1} \times 0.5 + 40 \times 9.8 \times 0.125 + N_{2} \times 0.5 = 0

(N_{1} - N_{2}) \times 0.5 = 40 \times 9.8 \times 0.125

N_{1} - N_{2} = 392 \times 0.125 \times 2 = 98

...(ii)

Add (i) and (ii),

2N_1 = 490

N_{1} = \frac{490}{2} = 245\,\text{N}

From (

i ), N_{2}=392-N_{1}

=392-245=147\,\text{N}

For rotational equilibrium of side AB of the step ladder, taking moments about A, we get

M g \times F H - N_{1} \times B K + T \times A G = 0

40 \times 9.8 \times 0.125 - 245 \times 0.5 + T \times 0.76 = 0

\therefore T \times 0.76 = 245 \times 0.5 - 40 \times 9.8 \times 0.125

= 122.5 - 49 = 73.5

T = \frac{73.5}{0.76} = 96.7\,\text{N}

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