# A typical compound archery bow requiers

• October 18th 2010, 08:07 AM
r-soy
A typical compound archery bow requiers
A typical compound archery bow requiers a force of 133 N to hold an arrow at "full drwa " ( pulled back 71 cm ). ssunimg that the bow obeys hooke's law . what is its spring constant ?

did conver 71 to m
did use k = W/x to find spring constant ?

plese help me ?
• October 19th 2010, 03:13 AM
HallsofIvy
Yes, that's right. Hooke's law says that the force on a spring is given by the spring constant times the distance the spring is extended: F= kx. So k= F/x or your W/x= 71 N/.71 m= 100 N/m.
• October 19th 2010, 04:30 AM
r-soy
Hi

F =W/x= 133N/.71 m= 1.87N/m.

why you make F =W/x= 71 N/.71 m = 100

??
• October 21st 2010, 03:13 AM
r-soy
plese I want your help .
• October 21st 2010, 05:15 AM
earboth
Quote:

Originally Posted by r-soy
Hi

F =W/x= 133N/.71 m= 1.87N/m. <== this is obviously wrong (Your arrow has a length of 71 m!)

why you make F =W/x= 71 N/.71 m = 100 <== this is obviously a typo

??

Quote:

Originally Posted by r-soy
plese I want your help .

The correct answer should be $k = 187\ \frac Nm$
• October 21st 2010, 10:48 AM
r-soy
187 X 133 /71 = 350.2

and what is the uint ??

and from where we got the 187 ?
• October 21st 2010, 01:19 PM
RHandford
Yes I agree that this is the correct answer. The unit is Newton/meter. 187 is the result that has been calculated i.e. 133/.71
• October 21st 2010, 10:50 PM
r-soy
I don't now from where we got 187 ??

on the book writen
K = mg/x , where k = force constant or spring constant
• October 21st 2010, 11:46 PM
earboth
Quote:

Originally Posted by r-soy
A typical compound archery bow requiers a force of 133 N to hold an arrow at "full drwa " ( pulled back 71 cm ). ssunimg that the bow obeys hooke's law . what is its spring constant ?

That's the question which should be answered. Nothing more, right?

Quote:

Originally Posted by HallsofIvy
... Hooke's law says that the force on a spring is given by the spring constant times the distance the spring is extended: F= kx. ...

In other words: You need a force F to extend the spring by a distance d. And you need a constant k which refers to the properties of the individual spring. Hooke's law states then:

$F=k \cdot d~\implies~k=\dfrac Fd$

F = 133 N
d = 0.71 m

you'll get: $k=\dfrac{133\ N}{0.71\ m}\approx 187\ \dfrac Nm$

Quote:

Originally Posted by r-soy
187 X 133 /71 = 350.2

...

What exactly do you want to calculate by that?
I'll include the units into this term and you'll get:

$187\ \frac NM \cdot \dfrac{133\ N}{71\ cm}$

The result has the "unit" $\dfrac{N^2}{cm \cdot m}$

That has definitely nothing to do with a compound bow.

Quote:

Originally Posted by r-soy
I don't now from where we got 187 ??

on the book writen
K = mg/x , where k = force constant or spring constant

This formula is only valid if a weight pulls at the spring (perpendicularly to the surface of Earth) and extend it.
• October 22nd 2010, 01:09 AM
r-soy
thank you

I have anothor quetion now I drwa graph between

http://www.mathhelpforum.com/math-he...n-dkm30822.jpg

How do we find k from graph

Are we calculate the slope then finid k

like = 51 - 19 / 0.2 - 0.1 = 32/0.1 but convert 32 to m then 0.032/0.1 = 0.32 but what id here the unit of solpe ?

and now find k = 1/0.032 =3.125 and what is the uint here ?

plese check mu answer is correct and what is the uint of slope and k ?