The figure shows a uniform, horizontal beam (length = 10 m, mass = 25 kg) that i

Question

The figure shows a uniform, horizontal beam (length = 10 m, mass = 25 kg) that is pivoted at the wall, with its far supported by a cable makes an angle of 51 degree with the horizontal. If a person (mass = 60 kg) stands 3.0 m from the pivot, what is the tension in the cable? 0.83 kN 0.30 kN 0.38 kN 0.42 kN 3.0 kN A uniform beam having a mass of 60 kg and a length of 2.8 m is held in place at its lower end by a pin. Its upper end leans against a vertical frictionless wall as shown in the figure. What is the magnitude of the force the pin exerts on the beam? 0.68 kN 0.57 kN 0.74 kN 0.63 kN 0.35 kN

Explanation / Answer

44) C)0.38 kN

let

L = 10m, m = 25 kg

theta = 51 degrees

M = 60 kg

let T is the tension in the cable.

As the beam is in equilibrium, net torque acting on the beam musbe zero.

Apply net torque on the beam about pivot = 0

3*M*g*sin(90) + m*g*(L/2)*sin(90) - T*L*sin(51) = 0

T = (3*M*g + m*g*(L/2) )/(L*sin(51))

= (3*60*9.8 + 25*9.8*5)/(10*sin(51))

= 384 N

= 0.384 kN

45) a) 0.68 kN

let Fx and Fy are the components of force exerted by pin
Apply, Fnety = 0

Fy - m*g = 0

Fy = m*g

= 60*9.8

= 588 N

Let Fw is the force exerted by wall on the rod

Apply net torque about = 0

FW*L*sin(40) - m*g*(L/2)*sin(50) = 0

FW = m*g*0.5*sin(50)/sin(40)

= 60*9.8*0.5*sin(50)/sin(40)

= 350 N

Apply, Fnet = 0

Fx - Fw = 0

Fx = Fw = 350 N

so, Fnet = sqrt(Fx^2 + Fy^2)

= sqrt(350^2 + 588^2)

= 684 N

= 0.684 kN

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