5. An 80 kg beam of length 4.00 m rests on two support points as shown in the fi

Question

5. An 80 kg beam of length 4.00 m rests on two support points as shown in the figure. The beam has a length of 4.00 m, and a uniformly distributed mass. Support point 1 is

1.50 m from the left end and support                       mmmmmmmmmmmmmmmmmmmmmmmm

                          1                                          2

point 2 is exactly on the right end.                                

a) What is the force that each support is applying to the beam?

b) A man of mass 60 kg wants to stand on the beam. At what distance from the left side of the beam must he stand so that each support is applying an equal amount of force to the beam?

c) The man begins walking to the left on the beam. At what distance from the left side of the beam is he standing when the beam begins to tip over?

Explanation / Answer

Let FL and FR are the forces exerted by left and right supports on beam.

Here, As the beam is in eqilibrium, net force and net torque acting on the beam must be zero.

Apply, Net torque about right support = 0

FL*2.5 - m*g*2 = 0

FL = 2*m*g/2.5

= 2*80*9.8/2.5

= 627.2 N

Apply Net torque about left support = 0

2.5*FR - 0.5*m*g = 0

FR = 0.5*m*g/2.5

= 0.5*80*9.8/2.5

= 156.8 N

b) here, FR = FL

Apply, Fnety = 0

FR + FY - 80*9.8 - 60*9.8 = 0

2*FR = 9.8*(80+60)

= 1372

FR = 686 N

let x is the distance from center(right side) where the person stand.

Apply Net torque about center = 0

-FL*0.5 - 60*9.8*x + FR*2 = 0

-686*0.5 - 60*9.8*x + 686*2 = 0

x = 1.5*686/(60*9.8)

= 1.75 m

so, from left end 2 + 1.75 = 3.75 m

c)

Apply, (1.5 - x)*60*9.8 - 80*9.8*0.5 = 0

1.5 -x = 80*9.8*0.5/(60*9.8)

1.5 - x = 0.667 m

x = 1.5 - 0.667

= 0.8333 m

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