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:
.
Physics - Damon, Wednesday, December 9, 2015 at 5:13am
F = k x
k = 2 g/6.1 cm
2.5g = (2g/6.1cm) x
x = 6.1 (2.5/2) cm
Answer:
a) F = 2250 Ib
b) F = 550 Ib
c) new max force ( F newmax ) = 2850 Ib
Explanation:
A) The force the wall of the elevator shaft exert on the motor if the elevator starts from rest and goes up
max capacity of elevator = 24000 Ibs
counterweight = 1000 Ibs
To calculate the force (F) :
we first calculate the Tension using this relationship
Counterweight (1000) - T = ( 1000 / g ) ( g/4 )
Hence T = 750 Ib
next determine F
750 + F - 2400 = 2400 / 4
hence F = 2250 Ib
B ) calculate Tension first
T - 1000 = ( 1000/g ) ( g/4)
T = 1250 Ib
F = 2400 -1250 - 2400/ 4
F = 550 Ib
C ) determine design limit
Max = 2400 * 1.2 = 2880 Ib
750 + new force - 2880 = 2880 / 4
new max force ( F newmax ) = 2850 Ib
(Direction) for the fact that it will continue having the momentum at the constant speed in which the engines turned off.
