You must observe the object twice.
-- Look at it the first time, and make a mark where it is.
-- After some time has passed, look at the object again, and
make another mark at the place where it is.
-- At your convenience, take out your ruler, and measure the
distance between the two marks.
What you'll have is the object's "displacement" during that period
of time ... the distance between the start-point and end-point.
Technically, you won't know the actual distance it has traveled
during that time, because you don't know the route it took.
Yes it is you are doing such a great job I just needed to tell u you will pass
Answer:
Explanation:
System of forces in balance
ΣFx = 0
ΣFy = 0
∑MA = 0
MA = F*d
Where:
∑MA : Algebraic sum of moments in the the point (A)
MA : moment in the point A ( N*m)
F : Force ( N)
d : Horizontal distance of the force to the point A ( N*m
Forces acting on the beam
T₁ = 620 N : Tension in cable 1 ,at angle of 30° with the vertical on the left
T₂ : Tension in cable 2, at angle of 50.0° with the vertical on the right.
W : Weight of the beam
x-y T₁ and T₂ components
T₁x= 620*sin30° = 310 N
T₁y= 620*cos30° = 536.94 N
T₂x= T₂*sin50°
T₂y= T₂*cos50°
Calculation of the T₂
ΣFx = 0
T₂x-T₁x = 0
T₂x=T₁x
T₂*sin50° = 310 N
T₂ = 310 N /sin50°
T₂ = 404.67 N
Calculation of the W
ΣFy = 0
T₂y+T₁y-W = 0
(404.67) *cos50° + 536.94 = W
W= 260.12+ 536.94
W= 797.06 N
Location of the center of gravity of the beam
∑MA = 0 , point (A) (point where the cable 2 of the right is located on the beam)
T₁y(5)-W(d) = 0
T₁y(5) = W(d)
d = T₁y(5)/W
d = 536.94(5) / 797.06
d = 3.37m
The center of gravity is located at 3.37m measured from the right end of the beam
