Answer:
Explanation:
According to question,
Charge 1 and charge 2 are
The distance between charges is 2 m
We need to find the force with which two positive charges repel. It is called electrostatic force of repulsion. It can be given by :
So, the electric force of repulsion is .
Answer:
the force P required for impending motion is 132.3 N
the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m
Explanation:
Given that:
mass of the cabinet m = 45 kg
coefficient of static friction μ = 0.30
A free flow body diagram illustrating what the question represents is attached in the file below;
The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.
Thus; considering the forces along the vertical axis ; we have :
The upward force and the downward force is :
where;
are the normal contact force at center point A and B respectively .
------- equation (1)
Considering the forces on the horizontal axis:
where ;
are the static friction at center point A and B respectively.
which can be written also as:
replacing our value from equation (1)
P = 0.30 ( 441)
P = 132.3 N
Thus; the force P required for impending motion is 132.3 N
b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm
the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at and it is shifted to
:
Then:
h = 0.8 m
Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m
