The system in the figure below is in equilibrium. A concrete block of mass 226 k

Question

The system in the figure below is in equilibrium. A concrete block of mass 226 kg hangs from the end of the uniform strut of mass 43.5 kg. A cable runs from the ground, over the top of the strut, and down to the block, holding the block in place. Angle = 30.0° and = 45.0°.

(a) Find the tension T in the cable.

(b) Find the horizontal component of the force on the strut from the hinge. (Include the sign with your answer. Assume that the positive direction is to the right.)

(c) Find the vertical component of the force on the strut from the hinge. (Include the sign with your answer. Assume that the positive direction is upward.)

The system in the figure below is in equilibrium. A concrete block of mass 226 kg hangs from the end of the uniform strut of mass 43.5 cable runs kg. A from the ground, over the top of the strut, and down to the block, holding the block in place. Angle 30.0 and 45.00. Strut Hinge (a) Find the tension T in the cable. (b) Find the horizontal component of the force on the strut from the hinge. (Include the sign with your answer. Assume that the positive direction is to the right.) (c) Find the vertical component of the force on the strut from the hinge. (Include the sign with your answer. Assume that the positive direction is upward.)

Explanation / Answer

Given,

M = 226 kg ; m = 43.5 kg ;

phi = 30 deg and theta = 45 deg

(a)We need to find tension T in the cable.

aplha = theta - phi = 45 - 30= 15 deg (this is the angle between cable and strut)

Its stated in the question that the system is in equilibrim, so the net torque on the system sums to zero. So,

-mg L/2 cos(theta) - Mg L cos(theta) + T sin(alpha) x L = 0

T sin(alpha) x L = L ( mg/2 cos(theta) + Mg cos(theta) )

L gets cancelled both the sides:

T = ( M + m/2) g cos(theta) / sin(alpha)

T = (226 + 43.5/2) x 9.81 x cos45 / sin15 = 6640.1 N

Hence, tension = T = 6640.1 N

b)The horizontal component will be:

Fx - T cos(phi) = 0

Fx = T cos(phi) = 6640.1 x cos30 = 5750.5 N

Hence, Fx = 5750.5 N

c)Vertical component will be:

Fy - mg - Mg - T sin(phi) =0

Fy = mg + Mg + T sin(phi)

Fy = 43.5 x 9.81 + 226 x 9.81 + 6640.1 x sin30 = 5961.15 N

Hence, Fy = 5961.15 N

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