Answer:
d)30,000 N
Explanation:
Solution:
mass(m)=1000kg
initial velocity (u)=0m/s
final velocity (v)=15m/s
So,
acceleration (a)=(v-u)/t
=(15-0)/10
=1.5m/s^2
Atlast,
Force(F)=m×a
=20000×1.5
=30,000 N
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
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To solve this problem, it is necessary to use the concepts related to the Force given in Newton's second law as well as the use of the kinematic equations of movement description. For this case I specifically use the acceleration as a function of speed and time.
Finally, we will describe the calculation of stress, as the Force produced on unit area.
By definition we know that the Force can be expressed as
F= ma
Where,
m= mass
a = Acceleration
The acceleration described as a function of speed is given by
Where,
Change in velocity
Change in time
The expression to find the stress can be defined as
Where,
F = Force
A = Cross-sectional Area
Our values are given as
Replacing at the values we have that the acceleration is
Therefore the force expected is
Finally the stress would be
Therefore the compressional stress that the arm withstands during the crash is 49.97Mpa