A crate with a mass of 175.5 kg is suspended from the end of a uniform boom. The

Question

A crate with a mass of 175.5 kg is suspended from the end of a uniform boom. The upper end of the boom is supported by the tension of 2886 N in a cable attached to the wall. The lower end of the boom pivots at the location marked X on the same wall. Calculate the mass of the boom.

A crate with a mass of 175.5 kg is suspended from the end of a uniform boom. The upper end of the boom is supported by the tension of 2886 N in a cable attached to the wall. The lower end of the boom pivots at the location marked X on the same wall. Calculate the mass of the boom. You'll need to get the various positions from the graph. Many are exactly on one of the tic marks.)

Explanation / Answer

m1 = 175.kg

T = 2886 N

let m2 is the mass and L is the length of boom.

let theta is the angle between T and Horizontal

tan(theta) = 2/11

theta = tan^-1(2/11) = 10.3 degrees

let alfa is the angle between boom and wall

tan(alfa) = 11/6

alfa = tan^-1(11/6) = 61.4 degrres

Net torque acting on the boom is zero.

T*L*sin(alfa+theta) - m2*g*(L/2)*sin(alfa) - m1*g*L*sin(alfa) = 0

T*sin(alfa+theta) - m2*g*sin(alfa)/2 - m1*g*sin(alfa) = 0

2886*sin(71.7) - m2*9.8*sin(61.4) - 175*9.8*sin(61.4) = 0

m2 = (2886*0.949 - 175*9.8*0.88)/(9.8*0.88) = 142.6 kg

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