Say you want to make a sling by swinging a mass M of 1.9 kg in a horizontal circle of radius 0.042 m, using a string of length 0
1 answer:
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Answer:
a)32.34 N/m
b)10cm
c)1.6 Hz
Explanation:
Let 'k' represent spring constant
'm' mass of the object= 330g =>0.33kg
a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.
ΣF=kx-mg=0
k=mg / x
k= (0.33 x 9.8)/ 0.1
k= 32.34 N/m
b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.
Therefore, amplitude of the oscillation is 10cm
c)frequency of the oscillation can be determined by,
f= 1/2π
f= 1/2π
f= 1.57
f≈ 1.6 Hz
Therefore, the frequency of the oscillation is 1.6 Hz
Complete Question
The complete question is shown on the uploaded image
Answer:
The tension on the shank is
Explanation:
From the question we are told that
The strain on the strain on the head is
The contact area is
Looking at the first diagram
At 600 MPa of stress
The strain is
At 450 MPa of stress
The strain is
To find the stress at we use the interpolation method
Substituting values
Generally the force on each head is mathematically represented as
Substituting values
Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as
