A 500.0-kilogram cannon is at rest on a frictionless surface. A remote triggering device causes a 5.0-kilogram projectile to be
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Answer:
= 0.66
Hence, the fraction of the length of the rod above water = = 0.66
and fraction of the length of the rod submerged in water = 1 - = 1 - 0.66 = 0.34
Explanation:
Data given:
Density of the rod = 5/9 of the density of the water.
Let's denote density of Water with w
And density of rod with r
So,
r = 5/9 x w
Required:
Fraction of the length of the rod above water.
Let's denote total length of the rod with L
and length of the rod above with = y
Let's denote the density of rod = r
And density of water = w
So, the required is:
Fraction of the length of the rod above water = y/L
y/L = ?
In order to find this, we first need to find out the all type of forces acting upon the rod.
We know that, a body will come to equilibrium if the net torque acting upon a body is zero.
As, we know
F = ma
Density = m/v
m = Density x volume
Volume = Area x length = X ( L-y)
So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:
F = mg
F = (Density x volume) x g
g = gravitational acceleration
F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)
where,
X (L-y) = volume
w = density of water.
Another force acting upon it is:
F = mg
F2 = X x L x r x g
Now, the torques acting upon the body:
T1 + T2 = 0
F1 ( y + ) g sinФ - F2 x (
) x gsinФ = 0
plug in the equations of F1 and F2 into the above equation and after simplification, we get:
=
. r
where, w is the density of water and r is the density of rod.
As we know that,
r = 5/9 x w
So,
=
. 5/9 x w
Hence,
=
=
Taking common and solving for
, we will get
= 0.66
Hence, the fraction of the length of the rod above water = = 0.66
and fraction of the length of the rod submerged in water = 1 - = 1 - 0.66 = 0.34