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?

Answers

Answer 1
Answer:

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

The given values are:

Weight,

  • mg = 500 N

Angle,

  • θ = 100°

As we know,

The sum of vertical forces = 0

then,

\Sigma Fy = 0

Now,

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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Just after a motorcycle rides off the end of a ramp and launches into the air, its engine is turning counterclockwise at 8325 rev/min. The motorcycle rider forgets to throttle back, so the engine's angular speed increases to 12125 rev/min. As a result, the rest of the motorcycle (including the rider) begins to rotate clockwise about the engine at 4.2 rev/min. Calculate the ratio IE/IM of the moment of inertia of the engine to the moment of inertia of the rest of the motorcycle (and the rider). Ignore torques due to gravity and air resistance.

Answers

Answer:

(Ie)/(lm) = 1.10*10^(-3)

Explanation:

GIVEN DATA:

Engine operating speed nf = 8325 rev/min

engine angular speed ni= 12125 rev/min

motorcycle angular speed N_m= - 4.2 rev/min

ratio of moment of inertia of engine to motorcycle is given as

(Ie)/(lm) = (-N)/((nf-ni))

(Ie)/(lm) = (-(-4.2))/((12125 - (8325)))

(Ie)/(lm) = 1.10*10^(-3)

Answer:1.105* 10^(-3)

Explanation:

Given

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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Answers

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Answers

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Explanation:

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