Law of motion and rope tension

[kring_c14]

Homework Statement

13) [20 points] Blocks A, B, and C are placed as in the figure and connected by ropes of
negligible mass. both “A” and “B” weigh 25.0 N each, the ramp-angle q = 36.9 and the coefficient of kinetic friction between each block and surface is 0.35. Block C descends with constant velocity.

b) Find the tension in the rope connecting blocks A and B.
c) What is the weight of block C?
d) If the rope connecting A and B were cut, what would be the acceleration of C?

for the figure pls see the attachment.

what I've worked out so far

the tension of A= 8.75 and the tension B= 8.66
how do i know which one is the answer for b.)

for the weight of block C i have no idea what to do..pls help...

 

[kring_c14]

by the way, how do you insert pictures without having to use attachment?/

 

[cristo]

Use [ /img] tags (without the space) if you've uploaded to an image sharing website, or just link to the site.

 

[kring_c14]

pls. help me with this one...pls. pls. pls..

 

[learningphysics]

Can you upload the image to sharing site? You can try this one:

http://www.imagevimage.com/

 

[kring_c14]

http://www.imagevimage.com/gallery.php?entry=images/sa.jpg

 

[kring_c14]

[url=http://www.imagevimage.com/gallery.php?entry=images/sa.jpg][PLAIN]http://www.imagevimage.com/thumbs/sa.jpg[/url][/PLAIN]

 

[kring_c14]

thanks a lot

 

[learningphysics]

8.75N looks right to me for part b)... how are you getting the 8.66N?

EDIT: you didn't post part a)

 

[kring_c14]

part a was all about drawing free body diagram... i thought i'll work it out by myself..

for 8.66
what i did is dis

i treat the objects separately
for tension A i got 8.75

for B:
$$\sum FxB$$ => Tcos36.9-fB=0
Tcos36.9= fB=$$\mu nB$$

$$\sum FyB$$ => Tsin36.9+nB-w=0
nB=w-Tsin36.9

solving the two eqn

T=$$\mu nB$$/cos36.9+$$\mu sin36.9$$

 

[learningphysics]

Part b) asks for the tension in the rope between A and B... that's 8.75N...

For the 8.66, you were using the rope between B and C right? So that wouldn't be what they want for part b... Though you do need this tension for the next parts... but 8.66 is the wrong value for the tension in this rope.

You chose your axes as x horizontal, y vertical... but the friction acts along the incline (not in the x - direction) And also, the normal force is not in the y direction...

Also, B has two tensions... the rope connecting to C, and the rope connecting to A, so you need to consider both forces in your freebody diagram.

I recommend choosing your axes differently to deal with the free body diagram of block B... take the x-axis along the incline... take the y-axis perpendicular to the incline... it'll make things easier...

 

[kring_c14]

im not sure if this is what you mean
T1 is the tension bet. rope A and B
T2 is bet B and C

im not really good at analyzing stuff

[url=http://www.imagevimage.com/][PLAIN]http://www.imagevimage.com/images/1_sa.jpg[/url][/PLAIN]

 

[learningphysics]

Yes, that's exactly what I mean...

Only thing is that the tension between A and B acting on B ie: T1... should also be along the x-axis... you only need to think about the rope beyond the pulley, right next to B...

 

[kring_c14]

the rope beyond the pulley? you mean the one in which block C is hanging?
so t1 and t2 should be both in the x axis..ahhhh get it! thanks

 

[learningphysics]

the rope beyond the pulley? you mean the one in which block C is hanging?
so t1 and t2 should be both in the x axis..ahhhh get it! thanks

Yeah, I mean the parts of the ropes attached to B... they are both along the x-axis.

 

[kring_c14]

how do i get the weight of C?

 

[kring_c14]

how bout the acceleration?/ my nosebleeds just staring at this problem...is it hard or am i justdumb...

 

[learningphysics]

You're not dumb at all... you're almost there... you made a small mistake:

T2 = T1 +wsin36.9 +$${\mu}N_b$$

That'll give you T2...

Then using the freebody diagram of C, you can get the weight of C...

 

[kring_c14]

wow thanks for lifting up my morale
I get cheesy when I feel sleepy...lol..and thanks again for helping me with this problem..

 

[learningphysics]

wow thanks for lifting up my morale
I get cheesy when I feel sleepy...lol..and thanks again for helping me with this problem..

lol. no prob.

 

[FedEx]

how bout the acceleration?/ my nosebleeds just staring at this problem...is it hard or am i justdumb...

You are not dumb. Never think like that. Now remenber one thing that while solving problem regarding Newtons laws of motion. The main thing is forming equations. Consider all sorts of forces which take part and relate them.Bingo! You get the answer

 

[truthlies217]

same question .. how can we get the weight of C?

 

[tiny-tim]

hi truthlies217!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[truthlies217]

mmm there are 3 forces applied on C all of them on Y axis .. F + n - mg = 0.0
we need mg so mg=F+n
where F = (FA+FB) >> I think not sure !
C must be heavier than 50 N to be able to pull the other two blocks ( my logical thinking ><) that's all what i got ><

 

[tiny-tim]

mmm there are 3 forces applied on C all of them on Y axis .. F + n - mg = 0.0

no, there are only two forces on C … mg and T (the tension)

there is no N: N signifies a normal reaction force, and a https://www.physicsforums.com/library.php?do=view_item&itemid=73" needs two surfaces in contact, but C isn't in contact with any other surface

call the other tension T₂, and do three https://www.physicsforums.com/library.php?do=view_item&itemid=26" equations (F = ma), one for each block …

what do you get? ​