If the tension in BC is 500N, determine the tension in BD and AB, and determine
the stretch in the spring.



Solution:


ΣFy=0\Sigma F_y=0
500 sin45FBDsin36.9=0500\ \sin45-F_{BD}\sin36.9=0
FBD=589NF_{BD}=589N

Fs=kΔlF_s=k\Delta l ΣFx=0\Sigma{F}_x=0
Δl=Fsk=825N800 N/m\Delta l=\frac{F_s}{k}=\frac{825N}{800\ N/m} 500cos45+589cos36.9=Fs500cos45+589cos36.9=F_s
Δl=1.03m\Delta l=1.03m Fs=825NF_s=825N
Determine the mass of A needed to hold the system at equilibrium.



Solution:
TBC=40g=392.4 NT_{BC}=40g=392.4\ N

ΣFy=0\Sigma{F}_y=0
392.4sin30TAE=0392.4sin30-T_{AE}=0
TAE=196.2NT_{AE}=196.2N
TAE=MAgT_{AE}=M_Ag
196.2N=MA(9.81m/S2)196.2N=M_A(9.81m/S2)
MA=20kgM_A=20kg
If l = 3ft, determine the tension each cable.

Solution:
2sinθ=3sin40  θ=25.37o\frac{2}{\sin\theta}=\frac{3}{\sin40}\ \rightarrow\ \theta=25.37^o

ΣFx=0\Sigma {F}_x=0
TABcos40TACcos25.37=0T_{AB}cos40-T_{AC}cos25.37=0
TAC=TABcos40cos25.37  (1)T_{AC}=T_{AB}\frac{\cos40}{\cos25.37}\ \ \left(1\right)

ΣFy=0\Sigma F_y=0
TAC sin25.37+TABsin40200=0T_{AC}\ \sin25.37+T_{AB}\sin40-200=0
(cos40cos25.37sin25.3+sin40)TAB=200\left(\frac{\cos40}{\cos25.37}\sin25.3+\sin40\right)T_{AB}=200
TAB=198.8NT_{AB}=198.8N
Sub into (1)

TAC=198.8cos40cos25.37T_{AC}=198.8\frac{cos40}{cos25.37}
TAC=168.5NT_{AC}=168.5N
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1) The lamp weights 57 N. Determine the tension in each of AB and AC needed to carry the lamp.



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2) Determine the resultant force on the hook:




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Determine the maximum allowable weight for the lamp at F if each of the cables can only handle a load of 50 N.





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Determine the tension in each of the cables supporting the phone tower under 7000 lb of compression.





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Determine the maximum allowable load if each of the cables can only handle 6 kN of tension.


Quiz: Practice Questions - Particle Equilibrium (Part 1)
Determine the maximum weight of the load if each cable can handle a tension of 100N.



Quiz: Practice Questions - Particle Equilibrium (Part 2)
Find the resultant force on the hook.

Determine the values of F1, F2 and F3 needed to keep the system at equilibrium.