BF2S Forums Physics Help.

haffeysucks wrote:

TheunforgivenII wrote:

3) A camper hangs a 26 kg pack between two trees, using two separate pieces of rope of different lengths. What is the tension in each rope?
(a) What is the tension in the left section of the rope (answer in N)
(b) What is the tension in the right section of the rope (answer in N)
On "a" I got 134.740 N and on "b" 144.289 N but both are wrong and I don't know why...read my textbook and followed the equation but got it wrong.

http://www.webassign.net/wp3/6-64.gif

the easiest way is to use matrices.  do you have a graphing calculator like the TI-84+?  it can solve it for you very easily.

i don't quite remember but you end up with two equations.

one was like T1 sin 71 + T2 sin 28 = mg
the other was, T1 cos 71 - T2 cos 28 = 0

this is how you enter it into the calculator.  under matrix A, make it 2x2.  make it sin 71, and sin 28.  under that, put cos 71, and -cos 28 (the negative is important!).
with matrix B, make it 2x1.  on top, put mg.  on the bottom, put 0.

now into the calculator type [A] ^-1 [b].  That little ^-1 is the inverse symbol, it's under the Math button.

at least, i think that's how it works.  that was last semester and i can't remember too well.

25.0, 8.04?

those two numbers you gave me look too small to be the answers

haffeysucks wrote:

TheunforgivenII wrote:

those two numbers you gave me look too small to be the answers

i just did it again and got 23.2 and 8.57.  i don't really know what's going on.  let someone else help you, i guess i can't remember this...

hm...well thanks for trying

mcminty wrote:

TheunforgivenII wrote:

3) A camper hangs a 26 kg pack between two trees, using two separate pieces of rope of different lengths. What is the tension in each rope?
(a) What is the tension in the left section of the rope (answer in N)
(b) What is the tension in the right section of the rope (answer in N)
On "a" I got 134.740 N and on "b" 144.289 N but both are wrong and I don't know why...read my textbook and followed the equation but got it wrong.

http://www.webassign.net/wp3/6-64.gif

For (a) I got 163.73N and for (b) I got 113.54N. How many sig.figures do you need? o.O

If they are right, I'll post my working when I'm online later.

I've tried 163.73 N on (a) and it was wrong and I just entered 113.54 N for (b) and it was wrong. I also tried 144.289 N and 271.368 N for (b) and still got it wrong. These numbers look really close together but I can't seem to get it.

mcminty wrote:

As the system is in equilibrium (ie there is no net force acting on the system), the sum of forces in both the X and Y directions has to be 0. I plugged the values I got back into the ∑Fx = 0 equation and the ∑Fy = 0. They both worked out.. so there is nothing wrong with what I have calculated.

I'm pretty sure I'm right...


(I'm off to work now, I'll post my working when I get back for you to have a look at)

my homework is on WebAssign and when I enter the numbers in for the answer they said it was wrong

mcminty wrote:

TheunforgivenII wrote:

my homework is on WebAssign and when I enter the numbers in for the answer they said it was wrong

What are you using for the gravitational acceleration constant?

Also, what was the equation that your textbook gave... you may have put the numbers in wrong (tbh, you should have posted that with the question, so we can see the methodology that your webassign expects).

There wasn't really an equation in the textbook, it's about Newton's Second Law in two dimensions under influence of multiple forces and also force and motion equations. The chapter in the book just told us to draw a free-body diagram and form an equation from the diagram. I'm going to try to do these problems myself first.

mcminty wrote:

TheunforgivenII wrote:

There wasn't really an equation in the textbook, it's about Newton's Second Law in two dimensions under influence of multiple forces and also force and motion equations. The chapter in the book just told us to draw a free-body diagram and form an equation from the diagram. I'm going to try to do these problems myself first.

Hmm..

This is a simple mechanics (statics) question. I redid it. I actually fucked up the diagram to determine the force components. They are NOT exact, but I got a ballpark figure of (a) 228N and (b) 85N.


Still... what does the book tell you to use for acceleration due to gravity? I'm using 9.81m/s/s.. because that's what we use. But I know in high school I used only 9.8.

sig figs maybe?

Anyway, this is how I did it...



EDIT: Putting the T1 and T2 values back into 2ii give a value of about 255.3N.. its close, but not "exact". Tho that's close enough for any practical engineering sense.

1) Two large crates, with masses 640 kg and 490 kg, are connected by a stiff, massless spring (k = 7.5 kN/m) and propelled along an essentially frictionless, level factory floor by a force applied horizontally to the more massive crate. If the spring compresses 4.9 cm from its equilibrium length, what is the applied force?

~okay I drew a free-body diagram and labeled everything on there and came up with the F = kΔx(m₁/m₂ + 1) and plugged in all the numbers which looked like this: F = 7.5 kN/m(-4.9)((490/640) + 1) = -64.88 N. This answer was wrong, so then I converted 7.5 kN/m to 7500 N/m. Re-entered all the numbers again and came up with -64886.71875 N, but that number looks strange to be the answer.....so now i'm stuck once again.

NooBesT wrote:

Convert cm to m too.

You should always include the units when you put the values in.

I did convert 4.9 cm to meters and re-did the equation to: F = 7500 N/m(-0.049m)((490/640) + 1 ) and got -648.867 N and was still wrong. I know I have a small mistake somewhere but I don't know where. When the spring is compressed does it mean negative?

TheunforgivenII wrote:

1) Two large crates, with masses 640 kg and 490 kg, are connected by a stiff, massless spring (k = 7.5 kN/m) and propelled along an essentially frictionless, level factory floor by a force applied horizontally to the more massive crate. If the spring compresses 4.9 cm from its equilibrium length, what is the applied force?

~okay I drew a free-body diagram and labeled everything on there and came up with the F = kΔx(m₁/m₂ + 1) and plugged in all the numbers which looked like this: F = 7.5 kN/m(-4.9)((490/640) + 1) = -64.88 N. This answer was wrong, so then I converted 7.5 kN/m to 7500 N/m. Re-entered all the numbers again and came up with -64886.71875 N, but that number looks strange to be the answer.....so now i'm stuck once again.

Can you post your FBD? I wanna see how you have done it..

mcminty wrote:

TheunforgivenII wrote:

and came up with the F = kΔx(m₁/m₂ + 1)

..How?

F_x = ma_x
= F - F₂ = m₁a_x
= F - F₂ = m₁/m₂F_s
= F - kΔx(m₁/m₂ + 1)

I started with F = ma and started to plug in the tensions and Hooke's law. I might have done the whole problem wrong, i don't know.

This is what my free-body diagram looks like:

1) A 72 N force is needed to slide a 42-kg box across a flat surface at a constant velocity. What is the coefficient of kinetic friction between the box and the floor? (Use g = 9.8 m/s²).

okay I need help on this one. I worked everything out but I don't know if this answer is correct or not. First I did 42kg(9.8 m/s²) = 411.6. Then I did (72/411.6) = .1749 is this answer right or wrong?