Answer:
Explanation:
Length if the bar is 1m=100cm
The tip of the bar serves as fulcrum
A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum
The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.
Using moment
Sum of the moments of all forces about any point in the plane must be zero.
Let take moment about the fulcrum
100×20-F×2=0
2000-2F=0
2F=2000
Then, F=1000N
The force acting in the crate lid is 1000N
Option D is correct
A. 60 miles
B. 5 hours
Unless you are looking for slope, in which case the answer is different
Answer:
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Explanation:
We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is
w² = mg d / I
In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow
d = L / 2
The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated
I = ¼ m r2 + ⅓ m L2
I = m (¼ r2 + ⅓ L2)
now let's use the concept of density to calculate the mass of the system
ρ = m / V
m = ρ V
the volume of a cylinder is
V = π r² L
m = ρ π r² L
let's substitute
w² = m g (L / 2) / m (¼ r² + ⅓ L²)
w² = g L / (½ r² + 2/3 L²)
L >> r
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
