In the figure shown, a 34.0 kg uniform square sign, of edge L =2.00m, is hung fr

Question

In the figure shown, a 34.0 kg uniform square sign, of edge L =2.00m, is hung from a horizontal rod of length d_h=3.00 m and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance d_v = 4.00 m above the point where the rod is hinged to the wall. What is the tension in the cable? What is the horizontal component of the force on the rod from the wall? (Include the sign. Take the positive direction to the to the right.) W hat is the vertical component of the force on the rod from the wall? (Include the sign. Take the positive direction to be up.)

Explanation / Answer

m =34 kg , x1 =1 m, x2 =3 m

= arctan (4/3) =53.13

from net torque about he hinge, e get

T = ((1/2)mgx1 +(1/2)mgx2)/x2sin(theta)

T = ((0.5*34*9.8*1) +(0.5*34*9.8*3))/(3*0.8)

T =277.7 N

(b) Net forec along horizontal direction

Fx = Tcos(theta) = 277.7*cos(53.13)

Fx = 166.6 N

(c) Net forec along vertical direction

Fy = mg - Tsin(theta)

Fy =(34*9.8) - (277.7*0.8)

Fy =111.04 N

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