Structural Mechanics 功課一問1
Structural Mechanics 功課一問!
- Audrey HepburnLv 71 decade agoFavorite Answer
a. Vector denotion is denoted by bold letters.
In the figure, the relationship of the three vectors have the following relationships: C = A + B
The vector sum of A and B can also be found by drawing two corresponding parallel vectors and joining together. This is known as parallelogram law.
b. The conditions for the static equilibrium for a two-dimensional free body are:
1. The resultant force acting on the free body is zero.
2. The resultant moment acting on the free body is zero.
1should be satisfied since by Newton's 1st law of motion, the body keepsits initial motion state under the absence of net force.
2 should be satisfied, otherwise there will be a rotational motion of the body.
Denote T1 and T2 be the tensions in cable AB and cable BC respectively.
Since the system is at equilibrium, so the net force in the system is zero.
Vertically: T1sin60* + 600sin15* = 500
Tension in AB, T1 = 398 N
Horizontally: T1cos60* + T2 = 600cos15*
398cos60* + T2 = 600cos15*
Tension in BC, T2 = 381 N
Denote O be the vertex of the triangle.
That is, OA = OB = 4 m
AB = 2.5 + 2.5 = 5 m
Triangle OAB is isosceles.
Since the system is at equilibrium, the resultant force acting in the system is zero.
Angle OAB = Angle OBA = cos-1(2.5 / 4) = 51.32*
Vertically: Tsin51.32* = W
Tsin51.32* = 120
Tension, T = 154 N
Horizontally, R = Tcos51.32*
R = 154cos51.32*
Reaction between two spheres, R = 96.1 NSource(s): Myself~~~