Answer:
W = (F1 - mg sin θ) L, W = -μ mg cos θ L
Explanation:
Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane
Y Axis
N - 
=
N = W_{y}
X axis
F1 - fr - Wₓ = 0
fr = F1 - Wₓ
Let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
We substitute
fr = F1 - W sin θ
Work is defined by
W = F .dx
W = F dx cos θ
The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1
W = -fr x
W = (F1 - mg sin θ) L
Another way to calculate is
fr = μ N
fr = μ W cos θ
the work is
W = -μ mg cos θ L
Answer:
Explanation:
Given a square Piece whose side is 12 inches
Now square pieces are cut from each corner to make it a open box
Suppose x is the length of square piece at each corner
then
base square has a length of 
Dimension of new box is 
Volume 
For maximum volume differentiate with respect to x we get
we get x=6 and 4 but at x=6 volume becomes zero therefore x=4 is valid
Answer:
I = I₀ + M(L/2)²
Explanation:
Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.
The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.
The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀
The distance between the two axes is L/2 (total length of the rod divided by 2
From the parallel axis theorem we have
I = I₀ + M(L/2)²
