. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
1 answer:
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Using Newton's Second Law, F = ma, where F is the net force
So the net force is:
F = (6kg)(4m/s^2) = 24N
Since you are applying a horizontal force of 30N, we can find the force of friction by the difference of the net force and the applied force.
30N-24N = 6N
So the net force is:
F = (6kg)(4m/s^2) = 24N
Since you are applying a horizontal force of 30N, we can find the force of friction by the difference of the net force and the applied force.
30N-24N = 6N
Here is the complete question
The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L = 20 mm below the centriod
The missing image which is the remaining part of this question is attached in the image below.
Answer:
The horizontal shear stress at point H is
Explanation:
Given that :
The internal shear force V = 80 kN = 80 × 10³ N
The moment of inertia = 64,900,000
The length = 20 mm below the centriod
The horizontal shear stress can be calculated by using the equation:
where;
Q = moment of area above or below the point H
b = thickness of the beam = 10 mm
From the centroid ;
Q =
Q =
Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³
Q = ( ( 700) × (55) + ( 3150 ) ( 97.5) ) mm³
Q = ( 38500 + 307125 ) mm³
Q = 345625 mm³
The horizontal shear stress at point H is
