Find Force at Point A & B to Support Rod

[Mike McCue]

A bar 10 ft long carries a weight of 20 lbs, 6 ft from the end. What force must be applied at each end to support the rod.

P1 6 | 4 P2
|____________________________________|
B | A C
20lb

 

[cristo]

Welcome to the forums. Please note that, for homework questions, you are required to show some work before we can help you. Also, in the future, please post in the relevant homework/coursework forum.

So, how do you think you'll approach this question?

 

[Astronuc]

Please post homework problems in the appropriate Homework forum - in this case Introductory Physics or perhaps Engineering.

Also, we ask that students show some effort in solving the equation before asking for help.

This appears to be a static beam problem where one must find the reaction forces using sum of forces = 0, and sum of moments = 0.

 

[Mike McCue]

Welcome to the forums. Please note that, for homework questions, you are required to show some work before we can help you. Also, in the future, please post in the relevant homework/coursework forum.

So, how do you think you'll approach this question?

AY 10(6)+10(4)=100 100/10= 10

 

[Mike McCue]

20(6)=120 120/10=12 B=12
20(4)=80 80/10=8 C=8

 

[Astronuc]

The greater force has to be applied at the shorter moment arm.

One knows that P1 + P2 = 20 lbs, since the net forces must be zero.

Then pick on end and determine the moments about that end. If one picks P1, then the 20 lbs is at 6 ft, and the moment is therefore 120 lb-ft.

The force P2 is at the other end of the beam, at 10 ft, so its moment is _________, which must equal the 120 lb-ft moment in order to maintain static equilibrium.

 

[Mike McCue]

The greater force has to be applied at the shorter moment arm.

One knows that P1 + P2 = 20 lbs, since the net forces must be zero.

Then pick on end and determine the moments about that end. If one picks P1, then the 20 lbs is at 6 ft, and the moment is therefore 120 lb-ft.

The force P2 is at the other end of the beam, at 10 ft, so its moment is _______, which must equal the 120 lb-ft moment in order to maintain static equilibrium.

moment at 10' is 40 lbs.

 

[Mike McCue]

moment at 10' is 200lbs

 

[Astronuc]

Well the moment of P2 at 10 ft is P2*10 and one has to solve for P2, by

0 = (20 lbs)(6 ft) - P2 (10 ft), which gives P2 = _________


The use P1 + P2 = 20 lbs.


The moments have to cancel to obtain static equilibrium.

 

[Mike McCue]

[QUOTE=Mike McCue;the moment at 10ft would be 200lbs.

 

[Mike McCue]

20lbs times 6ft=120 minus 10ft = 110lbs

 

[Mike McCue]

20lbs times 6ft=120 minus 10ft = 110lbs

I am some what confused Help!

 

[Mike McCue]

(20lb(6ft)-P2 (10ft)=0
120=10P2
120/10=P2
P2=12

 

[Mike McCue]

120-P2(10ft)=0
120=P2(10)
120/10=P2
12=P2

 

[Astronuc]

0 = (20 lbs)(6 ft) - P2 (10 ft), which gives P2 = 120 lb-ft / 10 ft or P2 = 12 lb, which one did correctly in the two preceding posts.

Then P1 + P2 = 20 lb = P1 + 12 lb, so P1 = 8 lb.

 

[Mike McCue]

120-P2(10ft)=0
120=P2(10)
120/10=P2
12=P2

P1=8lbs
P2=12lbs

Thanks for your help Astronuc.