As shown in 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 $40kg$ 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).
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The given situation can be shown as:
$N_B$ = Force exerted on the ladder by the floor point B $N_C$ = Force exerted on the ladder by the floor point C T = Tension in the rope BA = CA = 1.6m DE = 0. 5m BF = 1.2m Mass of the weight, m = 40kg Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H. $\Delta\text{ABI}$ and $\Delta\text{AIC}$ are similar $\therefore$ BI = IC Hence, I is the mid-point of BC. DE || BC BC = 2 × DE = 1m AF = BA - BF = 0.4m ....(i) D is the mid-point of AB. Hence, we can write, $\text{AD}=\Big(\frac{1}{2}\Big)\times\text{BA}=0.8\text{m}\ ...(\text{ii})$ Using equations (i) and (ii), we get, FE = 0.4m Hence, F is the mid-point of AD. FG || DH and F is the mid-point of AD. Hence, G will also be the mid-point of AH. $\Delta\text{AFG}$ and $\Delta\text{ADH}$ are similar $\therefore\ \frac{\text{FG}}{\text{DH}}=\frac{\text{AF}}{\text{AD}}$ $\frac{\text{FG}}{\text{DH}}=\frac{0.4}{0.8}=\frac{1}{2}$ $=\Big(\frac{1}{2}\Big)\times0.25=0.125\text{m}$ In $\Delta\text{ADH},$
$AH = (AD^2 - DH^2)^{1/2} = (0.8^2 - 0.25^2)^{1/2} = 0.76m$ For translational equilibrium of the ladder, the upward force should be equal to the downward force. $N_c + N_B = mg = 392 .....(iii)$ For rotational equilibrium of the ladder, the net moment about A is,$ -N_B \times BI + mg \times FG + N_C \times CI + T \times $
$AG - T \times AG = 0 -N_B \times 0.5 + 40 \times 9.8 \times 0.125 + N_C \times 0.5 = 0 (N_C - N_B) \times 0.5 = 49 N_C - N_B = 98 ....(iv)$ Adding equations (iii) and (iv), we get, $N_C = 245N N_B = 147N$ For rotational equilibrium of the side AB, consider the moment about A. $-N_B \times BI + mg \times FG + T \times AG = 0 -245 \times 0.5 + 40 \times 9.8 \times 0.125 + T \times 0.76 = 0 T = 96.7N$
$N_B$ = Force exerted on the ladder by the floor point B $N_C$ = Force exerted on the ladder by the floor point C T = Tension in the rope BA = CA = 1.6m DE = 0. 5m BF = 1.2m Mass of the weight, m = 40kg Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H. $\Delta\text{ABI}$ and $\Delta\text{AIC}$ are similar $\therefore$ BI = IC Hence, I is the mid-point of BC. DE || BC BC = 2 × DE = 1m AF = BA - BF = 0.4m ....(i) D is the mid-point of AB. Hence, we can write, $\text{AD}=\Big(\frac{1}{2}\Big)\times\text{BA}=0.8\text{m}\ ...(\text{ii})$ Using equations (i) and (ii), we get, FE = 0.4m Hence, F is the mid-point of AD. FG || DH and F is the mid-point of AD. Hence, G will also be the mid-point of AH. $\Delta\text{AFG}$ and $\Delta\text{ADH}$ are similar $\therefore\ \frac{\text{FG}}{\text{DH}}=\frac{\text{AF}}{\text{AD}}$ $\frac{\text{FG}}{\text{DH}}=\frac{0.4}{0.8}=\frac{1}{2}$ $=\Big(\frac{1}{2}\Big)\times0.25=0.125\text{m}$ In $\Delta\text{ADH},$
$AH = (AD^2 - DH^2)^{1/2} = (0.8^2 - 0.25^2)^{1/2} = 0.76m$ For translational equilibrium of the ladder, the upward force should be equal to the downward force. $N_c + N_B = mg = 392 .....(iii)$ For rotational equilibrium of the ladder, the net moment about A is,$ -N_B \times BI + mg \times FG + N_C \times CI + T \times $
$AG - T \times AG = 0 -N_B \times 0.5 + 40 \times 9.8 \times 0.125 + N_C \times 0.5 = 0 (N_C - N_B) \times 0.5 = 49 N_C - N_B = 98 ....(iv)$ Adding equations (iii) and (iv), we get, $N_C = 245N N_B = 147N$ For rotational equilibrium of the side AB, consider the moment about A. $-N_B \times BI + mg \times FG + T \times AG = 0 -245 \times 0.5 + 40 \times 9.8 \times 0.125 + T \times 0.76 = 0 T = 96.7N$
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