Two wires are used to suspend a sign that weighs 500 N. The two wires make an angle of 100° between each other. If each wire is exerting an equal amount of force how much force does each wire exert?


Answer 1

The force each wire exert will be "390.62 N".

The given values are:


  • mg = 500 N


  • θ = 100°

As we know,

The sum of vertical forces = 0


\Sigma Fy = 0


F Cos (50) +F Cos (50) = mg

By substituting the value, we get

→                   2F Cos (50) = 500

→                                 F = (500)/(2 Cos (50))

→                                     = (500)/(2(0.64))

→                                     = 390.62 \ N  

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Engine operating speed nf = 8325 rev/min

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motorcycle angular speed N_m= - 4.2 rev/min

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Initial angular speed of engine(\omega _E)=8325 rpm

Final angular speed of engine(\omega _E_f)=12125 rpm

Initial angular speed of Motorcycle(\omega _M)=0 rpm

Final angular speed of engine(\omega _M_f)=4.2 rpm

as there is no external torque therefore angular momentum remains conserved

I_E\omega _E+I_M\omega _M=I_E\omega _E_f+I_M\omega _M_f

I_E\omega _E+=I_E\omega _E_f+I_M\omega _M_f

I_E\left ( \omega _E-\omega _E_f\right )=I_M\omega _M_f

(I_E)/(I_M)=(\omega _M_f)/(\omega _E-\omega _E_f)

(I_E)/(I_M)=(-4.2)/(8325-12125)=0.0011052\approx 1.105* 10^(-3)

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