Biomedical Engineering Reference
In-Depth Information
45°
5 kg
50 N + 50 N + (50 N)sin 45
¼
Vertical:
T
Y
¼
135.4 N
(50 N)cos 45
¼
Horizontal:
T
X
¼
35.4 N
T
2
T
Y
2
+
T
X
2
; hence,
T
¼
¼
139.9 N
75.3
.
sin
θ ¼
T
Y
/
T
¼
0.967. Hence, the angle of
T
with the horizontal is
θ ¼
5.7
Inclined Plane
In Example 1.6 of Chap.
1
, we analyzed the forces acting on a child on an inclined
plane. There we could see that the force necessary to maintain a child on a plane in
static equilibrium corresponds to a fraction of the weight of this child. More
exactly, this force of action is equal to the weight of the child, multiplied by the
sine of the angle of the inclination of the plane. Analyze Fig.
5.12
.
To transport an object from bottom to top using an inclined plane of height
h
,
the action force to be employed will be smaller, as the angle
of inclination is
smaller, but, in contrast, the length
d
of the ramp will be longer. This occurs because
the work done with or without a machine must be the same, if friction is neglected.
θ
