Help With Physics Please!!! - Topic

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Oct 29 2014 07:20pm

If I understood the problem correctly, this is what I've calculated:

First, let's check two start figures to solve this problem:

1)

betoggc

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#12

Oct 29 2014 07:22pm

Ok, so now let's write down forces. What forces we have? A man pulling down the rope and also we have the weight of his pack at the midpoint of the rope. That's it.

Now all we have to do is calculate the exact force that the man will have to put in the rope in order to have his pack stable at the midpoint of the rope by 1,7m and 0,15m.

Here we go:

1)

betoggc

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#13

Oct 29 2014 07:27pm

Alright.

Now, before start calculations, we have to check the cosine of angle "A" and "B":

(3,25)^2 + (1,7)^2 = a^2
10.5625 + 2,89 = a^2
13.4525 = a^2
√13.4525 = a
3.667766 = a

(3,25)^2 + (0,15)^2 = b^2
10.5625 + 0.0225 = b^2
10.585 = b^2
√10.585 = b^2
3,253459 = b

So,

1,7
---- = Cos (A)
a

1,7
------------ = Cos (A)
3,667766

0,463497 = Cos (A)

And

0,15
----- = Cos (B)
b

0,15
------------ = Cos (B)
3,253459

0,046104 = Cos (B)

Ok, now we have full conditions to solve this problem:

In the first question (1,7m), the system have to be stable, so If you sum all forces in horizontal and in vertical, the value must be zero. Don't worry about horizontal forces in this problem. All you have to care here are vertical forces.
Ok, so, to have a stable system, the forces "P" and "Ty" must be equal.

Ty = P

Ty = T . Cos(A)
P = m.g

(I will consider g = 10m/s^2 because makes calculations easier)

So, we will have this expression:
T.Cos(A) = m.g
T.0,463497 = 15.10
T.0,463497 = 150
T = 336,059 N

But the tension T is the same force F that the man is pulling down the rope. So T = F.
The force "F", in the first problem, is 336,059 N

Now let's do the same to the second problem (0,15m)

We will have again Ty = P, so:
T.Cos(B) = m.g
T.0,046104 = 15.10
T.0,046104 = 150
T = 3253,513 N

So, the force "F" in the second problem is 3253,513 N.

If I did these problems correctly, you will see that the force to put the pack by 0,15m is almost ten times bigger than the force to put the pack by 1,7m.

(sorry for english mistakes)

tradess

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#14

Oct 29 2014 08:33pm

Quote (Dontrunaway @ Oct 29 2014 07:30pm)

Should be...

a ) 158.74N (2 sig figs = 160N)
b ) 1596N (2 sig figs = 1600N)

If you need the work let me know.

Thank you very much. Also the work would be nice.

@Betoggc

Wow, the diagrams are nice, but your work is off.

Ill pay both of you.

thanks guys.

This post was edited by iCK on Oct 29 2014 08:35pm

Dontrunaway

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#15

Oct 29 2014 11:33pm

Not stated in my work, but T = Force of Person since you assume the branch the rope is on is a frictionless pulley since it is not stated otherwise.