The Earth’s orbit around the Sun is slightly elliptical. At Earth's closest approach to the Sun (perihelion) the orbital radiu
1 answer:
You might be interested in
Answer:
<em>1.228 x </em><em> mm </em>
<em></em>
Explanation:
diameter of aluminium bar D = 40 mm
diameter of hole d = 30 mm
compressive Load F = 180 kN = 180 x N
modulus of elasticity E = 85 GN/m^2 = 85 x Pa
length of bar L = 600 mm
length of hole = 100 mm
true length of bar = 600 - 100 = 500 mm
area of the bar A = =
= 1256.8 mm^2
area of hole a = =
= 549.85 mm^2
Total contraction of the bar =
total contraction =
==> = <em>1.228 x </em>
<em> mm </em>
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 = / T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.