Answer:
Solution is given in the attachments
A rocket takes off from Earth's surface, accelerating straight up at 63.2 m/s2. Calculate the normal force (in N) acting on an a
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Answer:
x ’= 1,735 m, measured from the far left
Explanation:
For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.
Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive
They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,
the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar
x_{cm} = 1.2 -1
x_ {cm} = 0.2 m
Σ τ = 0
w₁ 1.2 + mg 0.2 - W₂ x = 0
x =
x =
let's calculate
x = 2.9 1.2 + 4 0.2 / 8
x = 0.535 m
measured from the pivot point
measured from the far left is
x’= 1,2 + x
x'= 1.2 + 0.535
x ’= 1,735 m
Answer:
The maximum acceleration of the system is 359.970 centimeters per square second.
Explanation:
The motion of the mass-spring system is represented by the following formula:
Where:
- Position of the mass with respect to the equilibrium position, measured in centimeters.
- Amplitude of the mass-spring system, measured in centimeters.
- Angular frequency, measured in radians per second.
- Time, measured in seconds.
- Phase, measured in radians.
The acceleration experimented by the mass is obtained by deriving the position equation twice:
Where the maximum acceleration of the system is represented by .
The natural frequency of the mass-spring system is:
Where:
- Spring constant, measured in newtons per meter.
- Mass, measured in kilograms.
If and
, the natural frequency is:
Lastly, the maximum acceleration of the system is:
The maximum acceleration of the system is 359.970 centimeters per square second.