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Dorkman
Registered User regular

Been working on this seemingly simple statics/friction problem for quite some time and for the life of me I can't figure it out. Or rather, I can't seem to get the answer in the back of the book, so I can only assume I can't figure it out. Bad form, yeah yeah, but I don't have much else to go on.

Anywho, question and what I have thus far behind the spoiler!

Coefficient is .6, Find P for impending motion.

How I started was to find N, which I got an equation of:

**N=160.19 - 8/17 P, (Where 160.19 = 8/17(34.7 * 9.81))**

Which takes into account the perpendicular portion of W pushing downward and the perpendicular component of P relieving some of that weight.

From there I moved to the parallel forces, which are -Ff, 15/17 P, and 15/17W. I worked those into the formula of:

**.6(N) = 15/17 P + 300.359, (300.359 = 15/17(34.7 * 9.81))**

Substitute N with the the equation I found earlier to get:

**.6(160.19 - 8/17P) = 15/17 P + 300.359**

Work it down to 19.8/17 P = -204.245. Which doesn't seem right to have a negative force, as all the directions were pretty self explanatory.

Regardless that works down to P = -175.36, where the answer is 17.2. So yeah, I am messing up big somewhere I can't figure out where. I am sorry that my equations may be hard to read, but I couldn't think of an easier way to demonstrate this.

In my mind everything is where it should be and all the proper forces are taken into account.

(I left out units as having an N for Natural for and N's for Newtons would be just confusing.)

Help?

Anywho, question and what I have thus far behind the spoiler!

Coefficient is .6, Find P for impending motion.

How I started was to find N, which I got an equation of:

Which takes into account the perpendicular portion of W pushing downward and the perpendicular component of P relieving some of that weight.

From there I moved to the parallel forces, which are -Ff, 15/17 P, and 15/17W. I worked those into the formula of:

Substitute N with the the equation I found earlier to get:

Work it down to 19.8/17 P = -204.245. Which doesn't seem right to have a negative force, as all the directions were pretty self explanatory.

Regardless that works down to P = -175.36, where the answer is 17.2. So yeah, I am messing up big somewhere I can't figure out where. I am sorry that my equations may be hard to read, but I couldn't think of an easier way to demonstrate this.

In my mind everything is where it should be and all the proper forces are taken into account.

(I left out units as having an N for Natural for and N's for Newtons would be just confusing.)

Help?

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## Posts

Dance CommanderonIt's a 8/15 triangle (rise/run)

Dorkmanonbasically you have the angles wrong for the weight.

correct that and it'll probably work.

Dunadan019onI guess I should take this time to ask, is there any online resources to help myself figure out friction and static equilibrium? I have been doing well in this course up to this point, but for some reason friction has me absolutely baffled.

DorkmanonEdit: as far as looking for a resource, you can find 99% of everything you want to know before independent research / grad school on wikipedia.

http://en.wikipedia.org/wiki/Friction

rabidrabbitsonI do need further help. I hate making this like a "do my homework" type thread, but I am just really trying to grasp these concepts by Friday (Test day!).

The problem is, there is a 340mm high x 220 wide box on an inclined plane. No Weight is given, and no force of friction is given, however we are told that u = .7. The question is, at what point will this box begin to slide/tip, and determine if it will tip or slide.

My thinking was, draw a resultant from C.G. to the bottom corner of the box as it is inclined. In my head, the box will only tip if C.G. is no longer perfectly aligned with the bottom corner of the box (thinking of it in terms of a diamond type shape). With that resultant, I took the Tan of 110 / 170 (half the width divided by half the height.)

This makes sense to me, and I understand what I am doing. Furthermore it gives me the answer I am looking for. Sweet!

However, when I discussed this with my instructor, he did not agree, and accused me of getting the answer out of the back of the book and finding an equation that would make it work. Well poo.

In his mind, the proper way to find the answer would be to take a moment at the bottom corner of the box as it is being tipped. This would eliminate Ff, and N as they are at the same point when it is about to tip.

This leaves me with the formula:

W x cos A x 110mm = W x sin A x 170

Where W if the weight of the block, and A is the missing angle that we apparantly don't know. I notice this can be reduced to: 110CosA = 170SinA.

And..now I am stuck, where do I go from here? I thought for sure my original thinking was sound and made sense, but I guess not.

DorkmanonLemmingonSigh. Sometimes I wonder if I am cut out for the Mechanical Engineering Technology gig...I really do.

Dorkmanon