# Mechanics String and PUlley problems

• Oct 15th 2012, 09:58 AM
mysur
Mechanics String and PUlley problems
A mass of 10 kg lies on a smooth plane which is inclined at THETA to the horizontal. The mass is 5m from the top measured along the plane. One end ofa light inextensible string is attached to this mass; the string passes up the line of greatest slope and over a smooth pulley fixed at the top of the slope. The other end is attached to a freely suspended mass of 15 kg. This 15 kg mass is 4m above the floor. The system is released from rest and the string first goes slack 1 3/7 s later. Find the value of THETA?

• Oct 15th 2012, 10:30 AM
skeeter
Re: Mechanics String and PUlley problems
Quote:

Originally Posted by mysur
A mass of 10 kg lies on a smooth plane which is inclined at THETA to the horizontal. The mass is 5m from the top measured along the plane. One end ofa light inextensible string is attached to this mass; the string passes up the line of greatest slope and over a smooth pulley fixed at the top of the slope. The other end is attached to a freely suspended mass of 15 kg. This 15 kg mass is 4m above the floor. The system is released from rest and the string first goes slack 1 3/7 s later. Find the value of THETA?

we've done this before in another problem, no?

start by making a sketch, then by writing individual scalar net force equations for the hanging mass and the mass on the incline.
• Oct 15th 2012, 11:06 AM
MaxJasper
Re: Mechanics String and PUlley problems
Let me know if you get:

$\displaystyle \theta =30.067{}^{\circ}$ (Rock)
• Oct 15th 2012, 11:08 AM
mysur
Re: Mechanics String and PUlley problems
Thank you I did used the mg- mgsin theta = ma ( two masses)
15g - 10g sintheta= 25a
s= 4m a=? t= 1 3/7
s= 1/2 at ^2
4= 1/2 x a x (10/7)^2
a= 3.92 m/s^2

using 15g- 10gsin theta = 25a
where a = 3.92
15g- 10x9.8 sin theta= 25x3.92
we get sin theta= 15g-98/10g = 49/98= .5
theta= 30 degrees

can you please verify, Thank you very much
• Oct 15th 2012, 11:14 AM
mysur
Re: Mechanics String and PUlley problems
since it is theta to the horizontal it is cos theta is it
I m confused
• Oct 15th 2012, 11:30 AM
mysur
Re: Mechanics String and PUlley problems
Could you please verify and let me know where I went wrong?
• Oct 15th 2012, 11:49 AM
MaxJasper
Re: Mechanics String and PUlley problems
30 deg looks OK
• Oct 15th 2012, 12:18 PM
ebaines
Re: Mechanics String and PUlley problems
Quote:

Originally Posted by mysur
Thank you I did used the mg- mgsin theta = ma ( two masses)
... can you please verify, Thank you very much

Yes, this is correct.

Quote:

Originally Posted by mysur
since it is theta to the horizontal it is cos theta is it
I m confused

The force due to the weight of the 10Kg block that is opposing the falling of the 15 Kg weight is $\displaystyle mg \sin \theta$ just as you have it. I sometimes find it easiest to think through whether it's sine or cosine by imaging what happens if theta changes - for example if the slope was 0 degrees to the horizontal (perfectly level) then the weight of the 10Kg block doesn't present any force at all, and if theta = 90 then the full weight of the 10Kg mass would come into play. So it's pretty clear that you want to use sine theta in this problem.
• Oct 15th 2012, 01:12 PM
mysur
Re: Mechanics String and PUlley problems
Thank you for clairfying that