Lightweight, vertically suspended spiral spring with a spring constant of 8.6 N / m is fitted with 64 g weight. The weight shall
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Answer:
M₀ = 5i - 4j - k
Explanation:
Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e
M₀ = r x F
From the question,
r = i + j + k
F = 1i + 0j + 5k
Therefore,
M₀ = (i + j + k) x (1i + 0j + 5k)
M₀ =
M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)
M₀ = i(5) - j(4) + k(-1)
M₀ = 5i - 4j - k
Therefore, the moment about the origin O of the force F is
M₀ = 5i - 4j - k
Answer:
Explanation:
Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K
mg=k x
1 x 32= k x 2
K=16
And also give that damping force is 8 times the velocity so damping constant C=8.
We know that equation for spring mass system
my''+Cy'+Ky=F
Now by putting the values
1 y"+8 y'+ 16y=6 cos 4 t ----(1)
The general solution of equation Y=CF+IP
Lets assume that at steady state the equation of y will be
y(IP)=A cos 4t+ B sin 4t
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'= -4A sin 4t+4B cos 4t
y"=-16A cos 4t-16B sin 4t
Now put the values of y" , y' and y in equation 1
1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t
So by comparing the coefficient both sides
-16A+32B+16A=0 So B=0
-16 B-32 A+16B=6 So A=-3/16
y=-3/16 cos 4t
Now to find the CF of differential equation 1
y"+8 y'+ 16y=6 cos 4 t
Homogeneous version of above equation
So
So the general equation
Given that t=0 Y=0 So
t=0 Y'=0 So
The above equation is the general equation for motion.