Question

A beam having a length of 20 metres is pivoted at its mid point. A 200 newton load is located at a point 5 metres from the right hand end of the beam. A 300 newton load is located at a point 8 metres from the right hand end. In order to be in equilibrium, what load is required at the extreme left end of the beam?
F= Unknown

200N* 5m= 1000Nm
300N* 8m= 2400Nm
2400Nm + 1000Nm = 3400Nm
3400Nm/10m = 10m*F/10m
F= 340N
340N is required at the extreme left end of the beam to be in equilibrium.

Is this correct?

Answer 1

sum moments about the right end.

5*200+8*300+load*20-F*10=0 where F is the balance point force.

Summing vertical forces

F=300+200+load

5*200+8*300+load*20-(300+200+ load
)*10=0

3400+20Load-5,000-10Load=0

load=160N

I am not certain what you did.


Now, check. Sum moments about the left end. F is 660 at the balance point.

10(660)-12(300)-15*200=0
6660-3600-3000=0
0=0 checks

Answer 2

10(660)-12(300)-15*200=0
6600-3600-3000=0
0=0 checks

Answer 3

300N 200N
? |- 8m -|- 5m -
|_____________________________________|
R2 10M | 10m R1

A beam having a length of 20 metres is pivoted at its mid point. A 200 newton load is located at a point 5 metres from the right hand end of the beam. A 300 newton load is located at a point 8 metres from the right hand end. In order to be in equilibrium.
What load is required at the extreme left end of the beam?
Would you mind showing my the formulas used to determine the answer. I'm a little confused on a few things. Thank you.