Answer:
y= 240/901 cos 2t+ 8/901 sin 2t
Explanation:
To find mass m=weighs/g
m=8/32=0.25
To find the spring constant
Kx=mg (given that c=6 in and mg=8 lb)
K(0.5)=8 (6 in=0.5 ft)
K=16 lb/ft
We know that equation for spring mass system
my''+Cy'+Ky=F
now by putting the values
0.25 y"+0.25 y'+16 y=4 cos 20 t ----(1) (given that C=0.25 lb.s/ft)
Lets assume that at steady state the equation of y will be
y=A cos 2t+ B sin 2t
To find the constant A and B we have to compare this equation with equation 1.
Now find y' and y" (by differentiate with respect to t)
y'= -2A sin 2t+2B cos 2t
y"=-4A cos 2t-4B sin 2t
Now put the values of y" , y' and y in equation 1
0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t
So by comparing the coefficient both sides
30 A+ B=8
A-30 B=0
So we get
A=240/901 and B=8/901
So the steady state response
y= 240/901 cos 2t+ 8/901 sin 2t
Answer:
Yes
Explanation:
A body can possess velocity at the same time in horizontal and vertical direction
For example
A projectile
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

Where,
F = Force/Weight
A = Area
= Compression
= Original Length
According to the values given we have to




Replacing this values at our previous equation we have,



Therefore the Weight of the object is 3.82kN