# Greatest Force without causing the rod to slip

1. Nov 12, 2013

### frownifdown

1. Here is the problem http://i.imgur.com/DeCXHNE.png

2. F=-cos45-sin45μ+(weight*μ*cos45)/.5-cos45

3. I plugged the numbers in the equation which we got by combining other equations for F, and I got 9.96 and it isn't accepting it. My other friends used this equation with their problems and they got the right answers. I don't understand why mine isn't working

Last edited: Nov 12, 2013
2. Nov 12, 2013

### Staff: Mentor

Something missing from this post?

3. Nov 12, 2013

### frownifdown

Ah crap! I suppose it is. Fixed

4. Nov 12, 2013

### haruspex

Is that with all appropriate parentheses? Anyway, I don't get an equivalent equation. How did you arrive at the equation, what did number did you get, and what is the book's answer?

5. Nov 12, 2013

### frownifdown

To be honest, it was my buddy that got the equation. It was a long and complicated process but I know it is right because he got it without using numbers and it worked with his numbers and my other friends numbers, but for some reason it isn't working with mine.

Can anyone chip in with some help?

6. Nov 12, 2013

### haruspex

I asked two other questions: what number do you get and what is the correct number?
Btw, it doesn't seem very educational for you to be using your buddy's equation without knowing how it was derived. How about you post your own attempt to derive it?

7. Nov 12, 2013

### frownifdown

I know it isn't very educational but it isn't going to be on the test and I really don't understand how he did it. I got 9.56 as the answer but that isn't it, and we all did it with my numbers about three different times and got that each time. I don't know what the right answer is, or I wouldn't be asking here. I have like 4 hours to get this submitted and still have to study everything.

8. Nov 12, 2013

### haruspex

The equation you posted is flat wrong. It makes no sense since the answer must be proportional to the mass. I also suspect that you have omitted parentheses. It reads F=-cos45-sin45μ+(((weight*μ*cos(45))/.5)-cos45, which simplifies to F=-2cos45 - μ sin45 + 2*weight*μ*cos45. That calculates as 4.27, not 9.56 or 9.96.
I'm sorry you're under time pressure, but my aim (and the principle behind this forum) is to help you understand the physics, not merely pass tests.

9. Nov 12, 2013

### frownifdown

And I appreciate it greatly. It's so frustrating that I can't grasp these concepts and it's not something that I'm used to when it comes to math classes. So are you saying that the final answer is 4.27? Because it's not accepting that either.

10. Nov 12, 2013

### haruspex

No, I'm saying that (1) the equation as posted makes no sense and (2) it doesn't seem to be the equation you actually used since it doesn't give the number you say.
If you show some attempt to derive the equation I'll help you through it.

11. Nov 12, 2013

### frownifdown

Alright, well I'll give it my best. If you could just start me out though, I don't have my notes with me for the two equations that I started with.

12. Nov 12, 2013

### frownifdown

Alright I have only 18 minutes left for this. I know it's not useful for learning, but could someone just solve this for me? I will stick around and figure it out after, but I have to get this in. Difference between 80 and 95% on this HW.

13. Nov 12, 2013

### frownifdown

Alright well thanks for the help anyways

14. Nov 12, 2013

### haruspex

In the hope that you get back to this some time, let the tension in the cable be T. Write down the two standard ƩF=ma equations (a being 0 here), one for the vertical and one for the horizontal. Be careful with the normal force from the ground - do not assume it's mg. Remember that the tension in the cable is also pushing down on the rod.
You will need a third equation. That, of course, will come from taking moments about some point. I suggest using the point where the rod touches the ground; that will make the equation as simple as possible.