Forces

Daniel123 · 2007-10-09 05:42:24

An athlete with a broken leg is in traction while the bone is healing. The diagram shows a system of pulleys for providing the traction force. All the pulleys are frictionless so that the tension in the rope is the same everywhere. g = 9.8 N kg-¹

(sorry about he dodgy drawing)

a) find the magnitude of the total horizontal force exerted on the leg by the system
b) find the total upward force exerted on the leg by the system
c) explain why the force found in a) does not move the patient to the bottom of the bed.

I'm clueless. Thanks.

Last edited by Daniel123 (2007-10-09 10:06:29)

Daniel123 · 2007-10-09 09:48:39

No one?

Daniel123 · 2007-10-10 07:48:02

!

Any help at all?

Last edited by Daniel123 (2007-10-10 07:48:49)

JaneFairfax · 2007-10-10 10:16:33

Whats the answer at the back of the book? If you tell me, I might perhaps be able to work my way towards it.

Daniel123 · 2007-10-10 10:36:33

I think the book is a bit confused. It says:

a) tension T = 8g (yes, grams for some reason), T + T cos 40° = 138N

b) T + T cos 50° = 129N

c) The force in (a) is balanced by the frictional force between the patient and the bed.

I don't really see where the working for a) and b) comes from. :S

JaneFairfax · 2007-10-10 11:41:16

I see whats going on. There are three forces on the leg: the horizontal force, the vertical one and the one at 40°. The forces are all equal in magnitude, i.e. 8g newtons (g is the acceleration due to gravity, not grams). Hence the horizontal and the vertical forces are as the book says.

Daniel123 · 2007-10-11 03:15:42

I realised that. I'm an idiot.

Thanks.

Last edited by Daniel123 (2007-10-11 03:15:55)

Math Is Fun Forum

#1 2007-10-09 05:42:24

Forces

#2 2007-10-09 09:48:39

Re: Forces

#3 2007-10-10 07:48:02

Re: Forces

#4 2007-10-10 10:16:33

Re: Forces

#5 2007-10-10 10:36:33

Re: Forces

#6 2007-10-10 11:41:16

Re: Forces

#7 2007-10-11 03:15:42

Re: Forces

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