# Homework Help: Find force

1. Sep 20, 2007

### 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

Last edited: Sep 20, 2007
2. Sep 20, 2007

### cristo

Staff Emeritus
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?

3. Sep 20, 2007

### Astronuc

Staff Emeritus
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.

4. Sep 20, 2007

### Mike McCue

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

5. Sep 20, 2007

### Mike McCue

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

6. Sep 20, 2007

### Astronuc

Staff Emeritus
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.

7. Sep 20, 2007

### Mike McCue

moment at 10' is 40 lbs.

8. Sep 20, 2007

### Mike McCue

moment at 10' is 200lbs

9. Sep 20, 2007

### Astronuc

Staff Emeritus
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.

10. Sep 21, 2007

### Mike McCue

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

11. Sep 21, 2007

### Mike McCue

12. Sep 21, 2007

### Mike McCue

13. Sep 21, 2007

### Mike McCue

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

14. Sep 21, 2007

### Mike McCue

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

15. Sep 21, 2007

### Astronuc

Staff Emeritus
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.

16. Sep 21, 2007

P1=8lbs
P2=12lbs