In the figure below, a 37.0 kg uniform square sign, of edge L = 2.00 m, is hung

Question

In the figure below, a 37.0 kg uniform square sign, of edge L = 2.00 m, 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? N What are the magnitude and direction of the horizontal component of the force on the rod from the wall? (Include the sign. Take the positive direction to be to the right.) N What are the magnitude and direction of the vertical component of this force? (Include the sign. Take the positive direction to be upward.) N The system in the figure is in equilibrium, with the string in the center exactly horizontal. Block A weighs 35 N, block B weighs 55 N, and angle phi is 33 degree. Find tension T_1. N Find tension T_2. N Find tension T_3. N Find angle theta. o

Explanation / Answer

1)

a)

here

the angle between the cable and the rod is

theta = tan^-1(dv / dh)

theta = tan^-1(4/3)

theta = 53.13 deg

the torques equations to get the tension

T * sin(theta) * dh - m * g * (dh - L/2) = 0

T = m * g * (dh - L/2) / ( sin(theta) * dh)

T = ( 37 * 9.8 * ( 3 - 2/2)) / ( sin53.13 * 3 )

T = 302.167 N

b)

the x axis to the right and y axis upward . The balance of forces equations yield

X : Fx - T * cos(theta) = 0

Fx = T * cos(theta)

Fx = 302.167 * cos53.13

Fx = 181.3 N

c)

y:Fy + T * sin(theta) - m * g = 0

Fy = m * g - T * sin(theta)

Fy = 37 * 9.8 - 302.167 * sin(53.13)

Fy = 120.9 N

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