7.22 As shown in Fig.7.40, 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/s2)
(Hint: Consider the equilibrium of each side of the ladder separately.)
NB = force exerted on the ladder by the floor point B
NB = force exerted on the ladder by the floor point B
T = Tension in the rope
BA = CA = 1.6 m, DE = 0.5 m and BF = 1.2 m
Mass of the weight, m = 40 kg
The perpendicular drawn from point A on the floor BC, this intersects DE at mid-point H.
Δ ABI and ΔACI are congruent. Therefore BI = IC, I is the mid-point of BC. DE is parallel to BC.
BC = 2 x DE = 1 m and AF = BA – BF = 0.4 m…… (i)
D is the mid-point of AB, hence we can write
AD = (1/2) x BA = 0.8 ……. (ii)
Using equation (i) and (ii), we get FE = 0.4 m
Hence, F i
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The given situation can be shown as
NB = force exerted on the ladder by the floor point B
NB = force exerted on the ladder by the floor point B
T = Tension in the rope
BA = CA = 1.6 m, DE = 0.5 m and BF = 1.2 m
Mass of the weight, m = 40 kg
The perpendicular drawn from point A on the floor BC, this intersects DE at mid-point H.
Δ ABI and ΔACI are congruent. Therefore BI = IC, I is the mid-point of BC. DE is parallel to BC.
BC = 2 x DE = 1 m and AF = BA – BF = 0.4 m…… (i)
D is the mid-point of AB, hence we can write
AD = (1/2) x BA = 0.8 ……. (ii)
Using equation (i) and (ii), we get FE = 0.4 m
Hence, F is the mid-point of AD.
Hence G will also be the midpoint of AH.
ΔAFG and ΔADH are similar
FG = (1/2)DH = (1/2) X 0.25 = 0.125 M
In ΔADH, AH = - ) = – ) = 0.76 m
For translational equilibrium of the ladder, the upward force should be equal to the downward force, NC + NB = mg = 392 …… (iii)
For rotational equilibrium of the ladder, the net moment about A is
NB x BI +mg x FG + NC x CI + T x AG – T x AG = 0
NB x 0.5 +40 x g + 0.125 x NC x 0.5 = 0
(NC - NB) x 0.5 = 49
NC - NB = 98 ………. (iv)
Adding equation (iii) and (iv) we get
NC = 245 N, NB = 147 N
For rotational equilibrium of the side AB, considering the moment about A
NB x BI + mg x FG + T x AG = 0
245 x 0.5 + 40 x 9.8 + T x AG = 0
T = 96.7 N
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<p><picture><img src="https://images.shiksha.com/mediadata/images/articles/1729073966phpSbnSOn.jpeg" alt="" width="161" height="166"></picture></p><p>The given situation can be shown as</p><p>NB = force exerted on the ladder by the floor point B</p><p>NB = force exerted on the ladder by the floor point B</p><p>T = Tension in the rope</p><p>BA = CA = 1.6 m, DE = 0.5 m and BF = 1.2 m</p><p>Mass of the weight, m = 40 kg</p><p>The perpendicular drawn from point A on the floor BC, this intersects DE at mid-point H.</p><p>Δ ABI and ΔACI are congruent. Therefore BI = IC, I is the mid-point of BC. DE is parallel to BC.</p><p>BC = 2 x DE = 1 m and AF = BA – BF = 0.4 m…… (i)</p><p>D is the mid-point of AB, hence we can write</p><p>AD = (1/2) x BA = 0.8 ……. (ii)</p><p>Using equation (i) and (ii), we get FE = 0.4 m</p><p>Hence, F is the mid-point of AD.</p><p>Hence G will also be the midpoint of AH.</p><p>ΔAFG and ΔADH are similar</p><p>FG = (1/2)DH = (1/2) X 0.25 = 0.125 M</p><p>In ΔADH, AH = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>3</mn><mi>A</mi><mi>B</mi></mrow></mrow><mrow><mrow><mn>4</mn><mi>g</mi><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>? </mo><mrow><mi>θ</mi></mrow></mrow></mrow></mrow></mfrac><mo>)</mo></math></span> - <span title="Click to copy mathml"><math><mo>√</mo><mfrac><mrow><mrow><mn>11.46</mn></mrow></mrow><mrow><mrow><mn>19.6</mn></mrow></mrow></mfrac></math></span> ) = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>F</mi><mi>G</mi></mrow></mrow><mrow><mrow><mi>D</mi><mi>H</mi></mrow></mrow></mfrac><mo>=</mo><mi></mi><mfrac><mrow><mrow><mi>A</mi><mi>F</mi></mrow></mrow><mrow><mrow><mi>A</mi><mi>D</mi></mrow></mrow></mfrac><mo>=</mo><mi></mi><mfrac><mrow><mrow><mi>F</mi><mi>G</mi></mrow></mrow><mrow><mrow><mi>D</mi><mi>F</mi></mrow></mrow></mfrac><mo>=</mo><mi></mi><mfrac><mrow><mrow><mn>0.4</mn></mrow></mrow><mrow><mrow><mn>0.8</mn></mrow></mrow></mfrac><mo>=</mo><mi></mi><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></math></span> – <span title="Click to copy mathml"><math><mo>√</mo><mo> (</mo><msup><mrow><mrow><mi>A</mi><mi>D</mi></mrow></mrow><mrow><mrow><mn>2</mn><mi></mi></mrow></mrow></msup></math></span> ) = 0.76 m</p><p>For translational equilibrium of the ladder, the upward force should be equal to the downward force, NC + NB = mg = 392 …… (iii)</p><p>For rotational equilibrium of the ladder, the net moment about A is</p><p><span title="Click to copy mathml"><math><msup><mrow><mrow><mi>D</mi><mi>H</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span> NB x BI +mg x FG + NC x CI + T x AG – T x AG = 0</p><p><span title="Click to copy mathml"><math><mo>√</mo><mo> {</mo><mo> (</mo><msup><mrow><mrow><mn>0.8</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span> NB x 0.5 +40 x g + 0.125 x NC x 0.5 = 0</p><p> (NC - NB) x 0.5 = 49</p><p>NC - NB = 98 ………. (iv)</p><p>Adding equation (iii) and (iv) we get</p><p>NC = 245 N, NB = 147 N</p><p>For rotational equilibrium of the side AB, considering the moment about A</p><p><span title="Click to copy mathml"><math><msup><mrow><mrow><mn>0.25</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span> NB x BI + mg x FG + T x AG = 0</p><p><span title="Click to copy mathml"><math><mo>-</mo><mi></mi></math></span> 245 x 0.5 + 40 x 9.8 + T x AG = 0</p><p>T = 96.7 N</p>
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