Answer:
The force required to begin to lift the pole from the end 'A' is 240 N
Explanation:
The given parameters for the pole AB are;
The length of the pole, l = 10.0 m
The weight of the pole, W = 600 N ↓
The distance of the center of gravity of the pole from the side 'A' = 4.0 m
Let '
' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive
For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have
× 10.0 m - W × 4.0 m = 0
∴
× 10.0 m = W × 4.0 m = 600 N × 4.0 m
× 10.0 m = 600 N × 4.0 m
∴
= 600 N × 4.0 m/(10.0 m) = 240 N
The force required to begin to lift the pole from the end 'A',
= 240 N.
12 protons in the nucleus
Explanation:
first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s
now using the 3rd law of motion
v^2=u^2+2as
0=9+2a2
a= -9/4m/s^2
now force=mass×accelration
=2kg×(-9/4)m/s^2
=4.5 N
4.5 newton force applied on the book!
✌️:)