|
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:
|
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.
|
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:
haha what grade are you in, I am doing the exact same thing in my physics class.
|
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. 
|
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: haha what grade are you in, I am doing the exact same thing in my physics class.
Grade 11
|
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.
|
ma = Ft - Fellen*sin25 - Fbob*sin15 - Fg
If you assume the rope is lowering at a constant speed
Ft = Fellen*sin25 + Fbob*sin15 - Fg
|
United States24805 Posts
Send $5 to micronesia@paypal
edit: please post what you are still confused about, if anything
|
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
|
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
|
17082 Posts
Let's not get too complacent, now ;D
|
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: 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 
|
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: 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.
|
United States3824 Posts
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
|
|
|
|
|
|
EPT![[image loading]](http://img338.imageshack.us/img338/865/phyqb5.jpg)
![[image loading]](http://g.imageshack.us/img338/phyqb5.jpg/1/)
