- #1

ShawnD

Science Advisor

- 668

- 1

Here is the question

http://myfiles.dyndns.org/pictures/statics10-2.png

I'll just do the part where the wire is long to show you what I have problems with.

To answer the problem I drew a FBD for member AB. Here are my variables and what they mean:

T - the tension in the wire at angle z

By - force of friction pushing down on member AB at point B

Bx - horizontal reaction force pushing left on AB at point B

The only horizontal forces on the member are from T and Bx. I can say this:

Bx = Tcos(z)

Since the friction of By is 0.35 times the reaction force Bx, I can say this:

By = 0.35Bx

Since point A and B are equal distance from the centre of mass for member AB, and since the moments on AB must cancel out, I can say this:

By = Tsin(z)

2 equations are equal to By so I'll rewrite them:

0.35Bx = Tsin(z)

now substitute in once more:

0.35Tcos(z) = Tsin(z)

now factor out the T:

0.35cos(z) = sin(z)

now make it equal zero:

0 = sin(z) - 0.35cos(z)

put it into my calculator and I get this:

z = 19.29 degrees off the X axis

Ok now here comes the problem. I have attained an angle for my cable but how do I find how long my cable is? At the angle I have, I don't know the horizontal length of AB, I don't know the vertical, I don't know it's angle.

I've tried using the angle between AB and AC but that seems extremely hard; I can't even get that method started. I've tried changing the axies to line up with AB but that makes the By and Bx forces very tricky.

One thing I noticed about the forces on AB is that the vertical forces at A and B make a couple, and the horizontal forces at A and B make a couple. If I sum moments around the centre of mass It shows that the horizontal couple and the verticle couple are equal and opposite (they cancel out). Going that approach would

Is there some easy way to solve this problem that I am overlooking?

http://myfiles.dyndns.org/pictures/statics10-2.png

I'll just do the part where the wire is long to show you what I have problems with.

To answer the problem I drew a FBD for member AB. Here are my variables and what they mean:

T - the tension in the wire at angle z

By - force of friction pushing down on member AB at point B

Bx - horizontal reaction force pushing left on AB at point B

The only horizontal forces on the member are from T and Bx. I can say this:

Bx = Tcos(z)

Since the friction of By is 0.35 times the reaction force Bx, I can say this:

By = 0.35Bx

Since point A and B are equal distance from the centre of mass for member AB, and since the moments on AB must cancel out, I can say this:

By = Tsin(z)

2 equations are equal to By so I'll rewrite them:

0.35Bx = Tsin(z)

now substitute in once more:

0.35Tcos(z) = Tsin(z)

now factor out the T:

0.35cos(z) = sin(z)

now make it equal zero:

0 = sin(z) - 0.35cos(z)

put it into my calculator and I get this:

z = 19.29 degrees off the X axis

Ok now here comes the problem. I have attained an angle for my cable but how do I find how long my cable is? At the angle I have, I don't know the horizontal length of AB, I don't know the vertical, I don't know it's angle.

I've tried using the angle between AB and AC but that seems extremely hard; I can't even get that method started. I've tried changing the axies to line up with AB but that makes the By and Bx forces very tricky.

One thing I noticed about the forces on AB is that the vertical forces at A and B make a couple, and the horizontal forces at A and B make a couple. If I sum moments around the centre of mass It shows that the horizontal couple and the verticle couple are equal and opposite (they cancel out). Going that approach would

*probably*give me an answer in the end but the algebra would be very ugly.Is there some easy way to solve this problem that I am overlooking?

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