Greatest Force without causing the rod to slip

In summary, the problem presents a physics equation (F=-cos45-sin45μ+(weight*μ*cos45)/.5-cos45) and a calculation using this equation resulted in a number (9.96) that is not accepted. The equation was derived from combining other equations, and the individual asking for help does not understand why the equation is not working for their numbers. The answer should be around 4.27, but it is not being accepted either. The person asking for help is under time pressure and is struggling to understand the concepts. They are asking for help in solving the problem.
  • #1
frownifdown
72
0
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
 
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  • #2
frownifdown said:
1. Here is the problem



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

Something missing from this post?
 
  • #3
berkeman said:
Something missing from this post?

Ah crap! I suppose it is. Fixed :redface:
 
  • #4
frownifdown said:
2. F=-cos45-sin45μ+(weight*μ*cos45)/.5-cos45
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
haruspex said:
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?

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
frownifdown said:
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?
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
haruspex said:
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?

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
frownifdown said:
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.
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
haruspex said:
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.

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
frownifdown said:
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.
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
haruspex said:
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.

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
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
Alright well thanks for the help anyways
 
  • #14
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.
 

Related to Greatest Force without causing the rod to slip

What is the "Greatest Force without causing the rod to slip"?

The "Greatest Force without causing the rod to slip" is the maximum amount of force that can be applied to a rod without it slipping or losing its grip on another object.

Why is it important to know the "Greatest Force without causing the rod to slip"?

Knowing the "Greatest Force without causing the rod to slip" is important for ensuring the safety and stability of structures or objects that rely on rods to hold them in place. It also helps engineers and designers determine the appropriate materials and dimensions for rods in various applications.

What factors affect the "Greatest Force without causing the rod to slip"?

The "Greatest Force without causing the rod to slip" can be affected by various factors such as the weight and surface area of the object being held, the material and surface properties of the rod, and the angle of the force being applied.

How is the "Greatest Force without causing the rod to slip" calculated?

The "Greatest Force without causing the rod to slip" is calculated using the coefficient of friction between the rod and the object it is holding, as well as the angle of the force being applied. The formula for calculating this force is F = µ * N * cosθ, where µ is the coefficient of friction, N is the normal force, and θ is the angle of the force.

What are some practical applications of knowing the "Greatest Force without causing the rod to slip"?

Some practical applications of knowing the "Greatest Force without causing the rod to slip" include designing structures such as bridges, buildings, and cranes, as well as selecting appropriate materials for objects that require rods to hold them in place, such as shelves, curtain rods, and tent poles. It can also be useful in determining safe weight limits for objects being hung or suspended from rods.

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