A person bends over to grab a 20 kg object. The back muscle responsible for supporting his upper body weight and the object is l
ocated 2/3 of the way up his back (where it attaches to the spine) and makes an angle of 12 degrees with the spine. His upper body weighs 36 N. What is the tension in the back muscle? A. 1500 N
B. 1700 N
C. 2700 N
D. 3500 N
As per FBD of the given situation we can say that torque due to tension in the spine is counter balanced by the torque due to weight of upper part of the body and the weight of the object
The tension force is acting at an angle of 12 degree
while both weight are acting perpendicular to the length
If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:
where Fg is the gravitational attraction, that can be written as follows according Newton's Universal Law of Gravitation:
Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming we can treat both spheres as point charges), as follows:
since m₁ = m₂ = 0.0024 kg, and r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:
Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:
Since both charges are the same, the charge on each sphere is just the square root of (5):
Q = 2.066* 10⁻¹³ C.
2)
Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.
Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows: