A mass m1 = 229 kg from the end of a uniform strut which is held at an angle the

Question

A mass m1 = 229 kg from the end of a uniform strut which is held at an angle theta equals 43 degrees with respect to the horizontal. The cable supporting the strut is at angle alpha = 29.2 degrees with respect to the horizontal. The strut has a mass of 45.4 kg. Find the magnitude of the Tension, T, in the cable. Find the magnitude of the horizontal component of the force exerted on the strut by the hinge. Find the magnitude of the vertical component of the force exerted by the strut on the hinge.

Explanation / Answer

here the net torque = 0

torque due weight of strut(m2) = -m2*g*l/2*cos43

torque due to m1 = -m1*g*l*cos43

torque to T in rope = -T*sin29.2*l*cos43 + T*cos29.2*l*sin43 = T*sin(43-29.2)*l

net torque = -m1*g*l/2*cos43 -m2*g*l*cos43 + T*sin(43-29.2)*l = 0

-m2*g*l/2*cos43 - m1*g*l*cos43 + T*sin(43-29.2)*l = 0

-45.4*9.8*l/2*cos43 - 229*9.8*l*cos43 + T*sin(43-29.2)*l = 0

T = 7562.88 N

++++++++++++

along horizantal

-Tx + Fx = 0

Fx = Tx = T*cos29.2 = 7562.88*cos29.2 = 6601.8 N

along vertical

Ty - Fy = (m1+m2)*g

Fy = (229+45.4)*9.8 + (7562.88*sin29.2)

Fy = 6378.7 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.