Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
A projectile motion is characterized by motion moving in a direction of an arc. It is acted upon by two component vectors: the horizontal and vertical. These two vectors are independent of each other when it comes to time of flight. The horizontal direction travels at constant speed, while the vertical direction travels at constant acceleration due to gravity, The time for an object to reach the ground would be equal, whether dropped from the sampe point or thrown in a projectile motion. Of course, this is assuming ideality wherein there is no air resistance.
So, the hang up time, or the time the object stayed on air is calculated using this equation:
a = Δv/t
Δv is the change in velocity which is the initial velocity when it was dropped to when it reaches zero velocity when it hits the ground.
9.81 m/s² = |(0 - 7.3)|/t
t = 0.744 seconds
If both particles have the SAME electrical charge, then they repel.
If they have DIFFERENT electrical charge, then they attract.
Protons have + charge .
Electrons have - charge .
So two protons (A) or two electrons (D) push apart.
One proton and one electron (C) pull together.
