The man is trying to push the homogeneous 20-kg ladder AB up a wall by applying the horizontal force P. Determine the smallest value of Pthat would move the ladder. The coefficient of static friction between the ladder and both contact surfaces is 0.3. h 1.5m 2m Note that h = 5 m

Given Data: The mass of the ladder is m = 20 kg. The Horizontal force is P. The coefficient of static friction between the ladder and both contact surface is μ = 0.3. The free body diagram of the system is given by, The expression for the inclination of the ladder is given by, θ=tan-1hb Here, b is the distance between the base of the wall and point A. Substitute all the known values in above equation, θ=tan-152θ=68.198° The expression for the length of the ladder is given by, sinθ=hl Here, l is the length of the ladder. Substitute all the known values in above equation, sin68.198°=5 mll=5.387 m Since. the mass of ladder is concentrated at the centre of gravity of the ladder. Thus, the location of centre of gravity of ladder is half of its length. The expression for the centre of gravity of ladder is given by, l'=l2 Substitute all the known values in above equation, l'=5.387 m2l'=2.69 m The expression for the horizontal distance between the centre of gravity of the ladder and point A is given by, cosθ=b'l' Here, b' is the horizontal distance between the centre of gravity of the ladder and point A. Substitute all the known values in above equation, cos68.198°=b'2.69 mb'=1 m The expression for the distance of centre of gravity of ladder from the base of the wall is given by, x=b-b' Substitute all the known values in above equation, x=2 m-1 mx=1 m The expression for the friction force at point B is given by, fB=μRB Here, RB is the normal reaction at point B. The expression for the equilibrium of vertical forces is given by, fB+mg=RA Here, RA is the normal reaction at point A. Substitute all the known values in above equation, μRB+mg=RA ⋯⋯⋯⋯⋯⋯1 The expression for the friction force at point A is given by, fA=μRA The expression for the equilibrium of horizontal forces is given by, RB+fA=P Substitute all the known values in above equation, RB +μRA=P ⋯⋯⋯⋯⋯⋯2 Substitute the value of equation (1) in equation (2), RB +μμRB+mg=P Substitute all the known values in above equation, RB +0.30.3RB+20 kg×9.81 m/s2=PRB +0.09RB+58.86 kg·m/s21 N 1 kg·m/s2=P1.09RB+58.86 N=P ……………………3The expression for the equilibrium of moment about the base of the wall is given by, ∑M=0RB×h+mg×x=P×y+NA×b Here, y is the height of force P from the base. Substitute all the known values in above equation, RB×5 m+20 kg×9.81 m/s2×1 m=P×1.5 m+0.3RB+20 kg×9.81 m/s2×2 m5RB+196.2 N·m=1.5P+0.6RB+392.4 N·m4.4RB-1.5P=196.2 N·m ………………4 On solving equation (3) and (4), RB=102.88 N,P=171 N Thus, the smallest value of P that would move the ladder is 171 N.

Answered: The man is trying to push the homogeneous 20-kg ladder AB up a wall by applying the horizontal force P. Determine the smallest value of Pthat would move the…

Elements Of Electromagnetics

7th Edition

ISBN: 9780190698614

Author: Sadiku, Matthew N. O.

Publisher: Oxford University Press

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The man is trying to push the homogeneous 20-kg ladder AB up a wall by
applying the horizontal force P. Determine the smallest value of Pthat would move
the ladder. The coefficient of static friction between the ladder and both contact
surfaces is 0.3.
1.5 m
2 m
Note that h = 5 m

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