A 10.00 kg block is placed at the top of a long frictionless inclined plane angled at 37.9 degrees relative to the horizontal. T
1 answer:
You might be interested in
Answer:
M₂ = M then L₂ = L
M₂> M then L₂ = \frac{M}{M_{2}} L
Explanation:
This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive
∑ τ = 0
The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂
The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.
M L + M₁ 0 - m₂ L₂ = 0
M L - m₂ L₂ = 0
L₂ = L
From this answer we have several possibilities
* if the two masses are equal then L₂ = L
* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L
Answer:
Approximately . (Assuming that
, and that the tabletop is level.)
Explanation:
Weight of the book:
.
If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, .
As a side note, the and
on this book are not equal- these two forces are equal in size but point in the opposite directions.
When the book is moving, the friction on it will be equal to
, the coefficient of kinetic friction, times
, the normal force that's acting on it.
That is:
.
Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that applied force. The net force on the book shall be:
.
Apply Newton's Second Law to find the acceleration of this book:
.