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[H]Physics

Blogs > YPang
Post a Reply
YPang
Profile Blog Joined April 2007
United States4024 Posts
Last Edited: 2008-10-30 01:26:19
October 30 2008 01:18 GMT
#1
A 500kg piano is being lowered into position by a crane while two people steady it with ropes pulling it to the sides. Bob's rope pulls to the left, 15 (degree) below horizontal. with 500N of tension. Ellen's rope pulls toward the right, 25(degree) below horizontal.

a). What tension must ellen maintain in her rope to keep the piano descending vertically?

Since i know the piano is in place and isn't moving to the left or right the Fnet(x) is 0. So 500cos(15)-Xcos(25)=0 |as a result i got 532.89N. However the answer in the back of the book is 530N. Probably sig fig.

b). What is the tension in the vertical main cable supporting the piano?

This part i have problem with. The 500 kg piano I assumed its MOVING towards. Fnet=ma. 500(9.8)-T1=500a... There's 2 variables, and i can't find the acceleration....

Edit:
[image loading]

[image loading]


sMi.Gladstone | BW: B high| SC2: gold T_T
SiegeTanksandBlueGoo
Profile Blog Joined December 2007
China685 Posts
October 30 2008 01:23 GMT
#2
ok, first of all, I assume that the problem assumes that the speed that the piano is lowered is constant. Constant speed = no acceleration (derive that shit) so no net force up or down. Thus Ft of up should be equal to the downward force. 500 N x Sin 15 + 530 N Sin 25 = Ft.
What does the scouter say about his macro level? It's Over 9000 minerals!
nAi.PrOtOsS
Profile Blog Joined August 2007
Canada784 Posts
October 30 2008 01:26 GMT
#3
On October 30 2008 10:18 YPang wrote:
A 500kg piano is being lowered into position by a crane while two people steady it with ropes pulling it to the sides. Bob's rope pulls to the left, 15 (degree) below horizontal. with 500N of tension. Ellen's rope pulls toward the right, 25(degree) below horizontal.

a). What tension must ellen maintain in her rope to keep the piano descending vertically?

Since i know the piano is in place and isn't moving to the left or right the Fnet(x) is 0. So 500cos(15)-Xcos(25)=0 |as a result i got 532.89N. However the answer in the back of the book is 530N. Probably sig fig.

b). What is the tension in the vertical main cable supporting the piano?

This part i have problem with. The 500 kg piano I assumed its MOVING towards. Fnet=ma. 500(9.8)-T1=500a... There's 2 variables, and i can't find the acceleration....

Edit:
[image loading]

[image loading]


haha what grade are you in, I am doing the exact same thing in my physics class.
YPang
Profile Blog Joined April 2007
United States4024 Posts
Last Edited: 2008-10-30 01:29:36
October 30 2008 01:28 GMT
#4
On October 30 2008 10:23 SiegeTanksandBlueGoo wrote:
ok, first of all, I assume that the problem assumes that the speed that the piano is lowered is constant. Constant speed = no acceleration (derive that shit) so no net force up or down. Thus Ft of up should be equal to the downward force. 500 N x Sin 15 + 530 N Sin 25 = Ft.


I tried that didn't work, the answer is 5300N. I don't know how to get it.

edit: epically waiting for micronesia's response.
sMi.Gladstone | BW: B high| SC2: gold T_T
YPang
Profile Blog Joined April 2007
United States4024 Posts
October 30 2008 01:28 GMT
#5
On October 30 2008 10:26 nAi.PrOtOsS wrote:
Show nested quote +
On October 30 2008 10:18 YPang wrote:
A 500kg piano is being lowered into position by a crane while two people steady it with ropes pulling it to the sides. Bob's rope pulls to the left, 15 (degree) below horizontal. with 500N of tension. Ellen's rope pulls toward the right, 25(degree) below horizontal.

a). What tension must ellen maintain in her rope to keep the piano descending vertically?

Since i know the piano is in place and isn't moving to the left or right the Fnet(x) is 0. So 500cos(15)-Xcos(25)=0 |as a result i got 532.89N. However the answer in the back of the book is 530N. Probably sig fig.

b). What is the tension in the vertical main cable supporting the piano?

This part i have problem with. The 500 kg piano I assumed its MOVING towards. Fnet=ma. 500(9.8)-T1=500a... There's 2 variables, and i can't find the acceleration....

Edit:
[image loading]

[image loading]


haha what grade are you in, I am doing the exact same thing in my physics class.


Grade 11
sMi.Gladstone | BW: B high| SC2: gold T_T
SiegeTanksandBlueGoo
Profile Blog Joined December 2007
China685 Posts
October 30 2008 01:37 GMT
#6
oh fuck me for being retarded. Add to that 500 kg x 9.8 for the force of gravity. xD

Your homework problems are so simple compared to mine. ><

This is what I get for nerdy asian schools.
What does the scouter say about his macro level? It's Over 9000 minerals!
Dead9
Profile Blog Joined February 2008
United States4725 Posts
Last Edited: 2008-10-30 01:40:03
October 30 2008 01:38 GMT
#7
ma = Ft - Fellen*sin25 - Fbob*sin15 - Fg
If you assume the rope is lowering at a constant speed
Ft = Fellen*sin25 + Fbob*sin15 - Fg
micronesia
Profile Blog Joined July 2006
United States24780 Posts
Last Edited: 2008-10-30 01:43:43
October 30 2008 01:42 GMT
#8
Send $5 to micronesia@paypal

edit: please post what you are still confused about, if anything
ModeratorThere are animal crackers for people and there are people crackers for animals.
YPang
Profile Blog Joined April 2007
United States4024 Posts
October 30 2008 01:46 GMT
#9
On October 30 2008 10:37 SiegeTanksandBlueGoo wrote:
oh fuck me for being retarded. Add to that 500 kg x 9.8 for the force of gravity. xD

Your homework problems are so simple compared to mine. ><

This is what I get for nerdy asian schools.


omg, thats what i forgot... >_<. I actually did what you suggested befoer i posted this problem, but i forgot to add the 500kg x 9.8. and i didn't know w hat went wrong.. xD
sMi.Gladstone | BW: B high| SC2: gold T_T
YPang
Profile Blog Joined April 2007
United States4024 Posts
October 30 2008 01:48 GMT
#10
thanks all... i get it now.. TL even helps me with my homework... I guess doing my HW with my computer infront of me isn't a TOTAL distraction at all... :D
sMi.Gladstone | BW: B high| SC2: gold T_T
Empyrean
Profile Blog Joined September 2004
17070 Posts
October 30 2008 01:55 GMT
#11
Let's not get too complacent, now ;D
Moderator
nAi.PrOtOsS
Profile Blog Joined August 2007
Canada784 Posts
October 30 2008 03:06 GMT
#12
On October 30 2008 10:28 YPang wrote:
Show nested quote +
On October 30 2008 10:26 nAi.PrOtOsS wrote:
On October 30 2008 10:18 YPang wrote:
A 500kg piano is being lowered into position by a crane while two people steady it with ropes pulling it to the sides. Bob's rope pulls to the left, 15 (degree) below horizontal. with 500N of tension. Ellen's rope pulls toward the right, 25(degree) below horizontal.

a). What tension must ellen maintain in her rope to keep the piano descending vertically?

Since i know the piano is in place and isn't moving to the left or right the Fnet(x) is 0. So 500cos(15)-Xcos(25)=0 |as a result i got 532.89N. However the answer in the back of the book is 530N. Probably sig fig.

b). What is the tension in the vertical main cable supporting the piano?

This part i have problem with. The 500 kg piano I assumed its MOVING towards. Fnet=ma. 500(9.8)-T1=500a... There's 2 variables, and i can't find the acceleration....

Edit:
[image loading]

[image loading]


haha what grade are you in, I am doing the exact same thing in my physics class.


Grade 11


grade 11, omg omg omg awww i feel bad now cause im in grade 12
YPang
Profile Blog Joined April 2007
United States4024 Posts
October 30 2008 03:08 GMT
#13
On October 30 2008 12:06 nAi.PrOtOsS wrote:
Show nested quote +
On October 30 2008 10:28 YPang wrote:
On October 30 2008 10:26 nAi.PrOtOsS wrote:
On October 30 2008 10:18 YPang wrote:
A 500kg piano is being lowered into position by a crane while two people steady it with ropes pulling it to the sides. Bob's rope pulls to the left, 15 (degree) below horizontal. with 500N of tension. Ellen's rope pulls toward the right, 25(degree) below horizontal.

a). What tension must ellen maintain in her rope to keep the piano descending vertically?

Since i know the piano is in place and isn't moving to the left or right the Fnet(x) is 0. So 500cos(15)-Xcos(25)=0 |as a result i got 532.89N. However the answer in the back of the book is 530N. Probably sig fig.

b). What is the tension in the vertical main cable supporting the piano?

This part i have problem with. The 500 kg piano I assumed its MOVING towards. Fnet=ma. 500(9.8)-T1=500a... There's 2 variables, and i can't find the acceleration....

Edit:
[image loading]

[image loading]


haha what grade are you in, I am doing the exact same thing in my physics class.


Grade 11


grade 11, omg omg omg awww i feel bad now cause im in grade 12


no don't worry, its a 12grade course... I'm not doing well in it if it makes you feel any better. XD.
sMi.Gladstone | BW: B high| SC2: gold T_T
cgrinker
Profile Blog Joined December 2007
United States3824 Posts
October 30 2008 03:11 GMT
#14
On October 30 2008 10:26 nAi.PrOtOsS wrote:
haha what grade are you in, I am doing the exact same thing in my physics class.


By the look of your drawing skills I would say 2nd
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