Answer:
Explanation:
The image attached to the question is shown in the first diagram below.
From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.
IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :
From the diagram; we can determine the length of BC by using pyhtagoras theorem;
SO;
[tex]L_{BC}^2 = L_{AB}^2 + L_{AC}^2[/tex]
[tex]L_{BC}^2 = (3.5+2.5)^2+ 4^2[/tex]
[tex]L_{BC}= \sqrt{(6)^2+ 4^2}[/tex]
[tex]L_{BC}= \sqrt{36+ 16}[/tex]
[tex]L_{BC}= \sqrt{52}[/tex]
[tex]L_{BC}= 7.2111 \ m[/tex]
The cross -sectional of the cable is calculated by the formula :
[tex]A = \dfrac{\pi}{4}d^2[/tex]
where d = 4mm
[tex]A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2[/tex]
A = 1.26 × 10⁻⁵ m²
However, looking at the maximum deflection in length [tex]\delta[/tex] ; we can calculate for the force [tex]F_{BC[/tex] by using the formula:
[tex]\delta = \dfrac{F_{BC}L_{BC}}{AE}[/tex]
[tex]F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}[/tex]
where ;
E = modulus elasticity
[tex]L_{BC}[/tex] = length of the cable
Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for [tex]L_{BC}[/tex] and 0.006 m for [tex]\delta[/tex] ; we have:
[tex]F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}[/tex]
[tex]F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN[/tex] ---- (1)
Similarly; we can determine the force [tex]F_{BC}[/tex] using the allowable maximum stress; we have the following relation,
[tex]\sigma = \dfrac{F_{BC}}{A}[/tex]
[tex]{F_{BC}}= {A}*\sigma[/tex]
where;
[tex]\sigma =[/tex] maximum allowable stress
Replacing 190 × 10⁶ Pa for [tex]\sigma[/tex] ; we have :
[tex]{F_{BC}}= 1.26*10^{-5} * 190*10^{6} \\ \\ {F_{BC}}=2394 \ N \\ \\ {F_{BC}}= 2.394 \ kN[/tex] ------ (2)
Comparing (1) and (2)
The magnitude of the force [tex]F_{BC} = 2.09676 \ kN[/tex] since the elongation of the cable should not exceed 6mm
Finally applying the moment equilibrium condition about point A
[tex]\sum M_A = 0[/tex]
[tex]3.5 P - (6) ( \dfrac{4}{7.2111}F_{BC}) = 0[/tex]
[tex]3.5 P - 3.328 F_{BC} = 0[/tex]
[tex]3.5 P = 3.328 F_{BC}[/tex]
[tex]3.5 P = 3.328 *2.09676 \ kN[/tex]
[tex]P =\dfrac{ 3.328 *2.09676 \ kN}{3.5 }[/tex]
P = 1.9937 kN
Hence; the maximum load P that can be applied is 1.9937 kN
