Answer:
γ
=0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ
= displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ
------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN
Answer:
The tension in the rope at the lowest point is 270 N
Explanation:
Given;
weight of the ball, W = 150 N
length of the rope, r = 4 m
velocity of the ball, v = 5.6 m/s
When the ball passes through the lowest point, the tension on the rope is the sum of weight of the ball and centripetal force.
T = W + F
Centripetal force, F = mv²/r
where;
m is the mass of the ball
m = W/g
m = 150 / 9.8 = 15.306 kg
Centripetal force, F = mv²/r
F = (15.306 x 5.6²)/4
F = 120 N
T = W + F
T = 150 + 120
T = 270 N
Therefore, the tension in the rope at the lowest point is 270 N
Answer:
I forget the word for it, but probably the guys who set up the power lines in the city.
Explanation: