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The formula for the center of mass (c.o.m) of a system of two parts is
_ m(1) X(1) + m(2) X(2)
(1) X = ---------------------
m(1) + m(2)
Where m(1) and m(2) are the masses of the parts and X(1) and X(2)
are the positions of the centers of mass of each of the constitutient
parts. Mow consider the figure below.
torso
-------O
|
Legs |
|_ c.o.m
.__|_____________ x-direction
| |
0cm 70cm
The mass of the torso ( call it m(2)) is 2/3 of 60kg
thus the mass of the legs ( call it m(1)) is 1/3 of 60kg.
>From the figure X(2) = 70 cm and X(1) = 0cm so we have
1 2
- 60kg x 0cm + - 60kg x 70cm
_ 3 3
(2) X = -------------------------------
1 2
- 60kg + - 60kg
3 3
_ 2800 kg cm __
(3) X = ------------ = 46.66 cm
60 kg
******************
* _ *
* X = 46.66 cm *
* *
******************
CATAGORY: Mechanics
QUESTION: The deltoid muscle raises the upper arm to a horizontal
position, as illustrated in the diagram. If the mass of the arm is 4.1 kg
and its centre of gravity is considered to act 36 cm from the shoulder
joint, (angle 18 degrees) find
a) the tension in the deltoid joint
b)the components Rx and Ry of the force exerted by the shoulder joint
c)the magnitude of the resultant force on the shoulder joint.
OK I think I have the picture here but unfortunately I do not have
complete information! You have not told me how far from the shoulder
joint the deltoid muscle attaches to the arm. Rather than guess I will
call this distance X (see figure below) and determine the$solution
in terms of X. OK now for the problem.
I see this as a hanging sign problem. See figure below. The angle a is
18 degrees. Gravity pulls the arm down with a force of
mg = 9.8m/s^2(4.1kg) = 40.18N
|
* Deltoid muscle connection
| *
| *
| *
shoulder a ( *
Joint--------------*-------------- arm
|_______X_______| |
| | |
|___________________36cm______|
| |
| |
| V F = mg = 40.18N
|
This is the old sum of the x and y forces must be equal to zero
type of problem. In the x-direction the muscle tension pulls the
bone against the shoulder joint and the shoulder joint responds
with a horizontal force Rx. If the muscle tension is T then
(1) x-direction forces
-T cos(a) + Rx = 0
In the y-direction the vertical part of the tension pulls up, the
shoulder pushes verticaly with Ry and gravity pulls down so
(2) T sin(a) + Ry - 40.18N = 0
We also have that the net torque about the shoulder joint must be
zero so
(3) X (T sin(a)) -36cm (40.18N) = 0
or X T sin(a) - 1446.48Ncm = 0
We will round 1446.48Ncm to 1450Ncm for convinence. We are mixing
the mks and cgs units here but we will be OK if we are careful to
think if X as always being in units of cm. We can solve (3) for T in
terms of X and a = 18 degrees. thus
(4) 1450Ncm
T = ------- = 4692.3 / X ~ 4690 / X
X sin(a)
We can now substitute (4) into (2) and solve for Ry
1450Ncm
(5) Ry = 40.18N - Tsin(a) = 40.18N - ------- sin(a)
X sin(a)
The sin(a) in the last term cancels giving
(6) Ry = 40.18N - 1450Ncm / X
Substituting (4) into (1) and solving for Rx gives
1450Ncm
(7) Rx = ------- cos(a)
X sin(a)
But cos(a)/sin(a) = cot(a) so
(8) Rx = 1450Ncm cot(a) / X
= 4462.6 / X
~ 4460 / X
The magnitude of the resultant force on the shoulder joint is then
(9) |R| = sqrt( Rx^2 +Ry^2)
Note: "sqrt" means square root and "^2" means squared
ie b^2 =b times b.
Thus in terms of X
(10) |R| = sqrt((4460 / X)^2 + (40.18N - 1450Ncm / X)^2)
OK lets write all the answers together in terms of the variable
X. X should be entered in the equations as centimeters.
************************************************************
* *
* (a) T = 4690Ncm / X *
* *
* (b) Rx = 4460Ncm / X *
* *
* Ry = 40.18N - 1450Ncm / X *
* *
* (c) |R| = sqrt((4460 / X)^2 + (40.18N - 1450Ncm / X)^2) *
* *
************************************************************
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