In Figure (a), a uniform 31.4 kg beam is centered over two rollers. Vertical lin

Question

In Figure (a), a uniform 31.4 kg beam is centered over two rollers. Vertical lines across the beam mark off equal lengths. Two of the lines are centered over the rollers; a 8.60 kg package of tamale is centered over roller B. What are the magnitudes of the forces on the beam from (a) roller A and (b) roller B? The beam is then rolled to the left until the right-hand end is centered over roller B (Figure (b)). What now are the magnitudes of the forces on the beam from (c) roller A and (d) roller B? Next, the beam is rolled to the right. Assume that it has a length of 1.01 m. (e) What horizontal distance between the package and roller B puts the beam on the verge of losing contact with roller A?

Explanation / Answer

given

M = 31.4 kg

L = 1.01 m

m = 8.60 kg

a )

the static equilibrium the normal force on beam from roller A is

FA = M g / 2

FA = 31.4 X 9.8 / 2

FA = 153.86 N

b )

for this here M g / 2 + m g

= 31.4 X 9.8 / 2 + 8.6 X 9.8

FB = 238.14 N

c )

same way the force is FB is

FA = M g + m g / 2

FA = ( M + m / 2 ) g

FA = ( 31.4 + 8.6 / 2 ) X 9.8

FA = 349.86 N

d )

FB = m g / 2

FB = 8.6 X 9.8 / 2

FB = 42.14 N

e )

m g x = M g ( L / 4 - x )

m x = M ( L / 4 - x )

8.6 x = 31.4 ( 1.01 / 4 - x )

8.6 x = 31.4 ( 0.2525 - x )

8.6 x + 0.2525 x = 7.9285

8.8525 x = 7.9285

x = 0.8956 m

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