A write 2.5 long with a cross-sectional area of 6mm^2. Stress den by 1.27mm when a mass of 45 kg is suspended from it. Find the stress on the arrival, the resulting strain of the value of young's modulus, Y of the wire
Answers
Answer:
Here we know that body explodes due to its internal forces
so there is no external force on it
and hence its COM will always move with same speed in same direction
so position of center of mass after t = 5 s is given as
\Delta x = 20(5) = 100 cmΔx=20(5)=100cm
x - 1 = 1x−1=1
x = 2
now we have
r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}r
cm
=
m
1
+m
2
m
1
r
1
+m
2
r
2
2 m = \frac{(2/3)(3, 2, -4) + (4/3)(x, y, z)}{2}2m=
2
(2/3)(3,2,−4)+(4/3)(x,y,z)
so we have
(4, 0, 0) = \frac{2}{3}(3, 2, -4) + \frac{4}{3}(x, y, z)(4,0,0)=
3
2
(3,2,−4)+
3
4
(x,y,z)
4 = 2 + \frac{4x}{3}4=2+
3
4x
x = \frac{3}{2}x=
2
3
0 = \frac{4}{3} + \frac{4}{3}y0=
3
4
+
3
4
y
y = -1y=−1
0 = -\frac{8}{3} + \frac{4}{3} y0=−
3
8
+
3
4
y
y = 2y=2
so position of heavy part is given as
r = (\frac{3}{2} , -1, 2)r=(
2
3
,−1,2)