Consider the situation below: 1.0 m 1.5 m -> ] Unknown mass A beam of mass 3 kg

Question

Consider the situation below: 1.0 m 1.5 m -> ] Unknown mass A beam of mass 3 kg and length 2.0 m is supporting a weight of unknown mass. The beam is supported by a cable which is attached to the wall 1.0 m above the beam and attached to the far end of the beam. It can sustain a maximum tension of 150 N. (a) What is the mass of the heaviest weight that can be suspended 1.5 meters from the wall? b) What are the horizontal and vertical components of the force by the pivot in this ca where the mass is what you determined in part (a))

Explanation / Answer

let the mass of unknown weight = m

M = mass of beam, L = length of beam

Tsin(theta) * L - mg *1.5 - Mg*L/2 = 0

150 * 1/sqrt(5) * 2 - m*9.8*1.5 - 3*9.8*1 = 0

134.16 - 14.7m - 29.4 = 0

m = 7.12 Kg

part b)

Horizontal component = Tcos(theta) = 150*2/sqrt(5) = 134.16 N

Vertical = Tsin(theta) - mg - Mg = 67.08 - 7.12*9.8 - 3*9.8 = 32.09 N (downwards)

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