3:39 PM 67%. ooooo T-Mobile LTE loncapa.tcc.fl.edu 19-5 For Due in 8 hours, Sale

Question

3:39 PM 67%. ooooo T-Mobile LTE loncapa.tcc.fl.edu 19-5 For Due in 8 hours, Sale Sign 19 minutes You are holding a sign as shown below. The sign (including the horizontal bar it hangs from) has a mass of 2.29 kg and is 60.0 cm wide. The sign is hanging from a 1.38 m tall, 5.14 kg vertical post. The sign is symmetric with a uniform mass distribution as implied by the drawing. FOR SALE What external force if up, if down) and torque for CCW, for CW) do you have to apply with your hand to keep the sign in static equilibrium? /Use g 9.80 m/s and ignore the width of the vertical post Applied force Applied torque: Submit Answer Tries 0/15

Explanation / Answer

Forces acting on the hand

1. weight of sign    = mg = 2.29*9.8 = 22.44 N

2. weight of the post - 5.15 *9.8 = 50.37 N

The force act vertically down = 22.44 + 50.37 = 72.81 N

The hand has to produces an equal and opposite force to keep it equilibrium

                                           = + 72.81 N (UP)

In addition to the force the sign produce a torque about the hand as it is away from the point of hand

torque by the sign about hand

          = F^ X r^

            = + 22.44 * 0.3 ( the sign is uniform and hence the weigth will acts at its CM and the troque is  CCW)

            = +6.732 N-m

we have to produce an opposite torque i.e. -6.732 N-m (CW)

-22

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