A13235378
- 50
- 10
- Homework Statement
- In a study on body balance, a physicist assembles an experimental apparatus, consisting of two bodies Q1 and Q2, of the same mass m, with charges + 2q and + q, respectively, separated by a distance of connected by strings of length l, to one with a thin, small stem of negligible mass. This stem is attached to a wire, which passes through four pulleys, the last of which is fixed to a rope that adheres to a block of mass M, which is on a surface with a friction coefficient µ. If the system does not reach equilibrium, the physicist adds plates to the body Q1 that generate a uniform electric field of value E, and to Q2 a magnetic field B, in addition to putting it in motion with a speed v, thus reaching , to static. However, when the wires that support the loads, by electrical repulsion, reach the same angle α with the vertical, the system is on the verge of movement
Consider:
- Gravity : g
- Middle electrical constant: K
- The threads are inextensible and of negligible mass
Which expression determines the value of the friction coefficient µ in the imminence of movement?
- Relevant Equations
- F=qvb
Fat=umg
Answer: E
My solution:
1) Load 2q:
Y direction:
$$ T_1cos\alpha=mg + 2Eq$$
2) Load q:
Radial direction:
$$ T_2-\frac{Fel}{sen\alpha}-\frac{Fmag}{cos\alpha}-\frac{mg}{cos\alpha}=\frac{mv^2}{l}$$
$$T_2cos\alpha=\frac{Fel}{tg\alpha}+Fmag+mg+\frac{mv^2cos\alpha}{l}$$
3) Stem connected to the wire
$$T=T'+T''=T_1cos\alpha+T_2cos\alpha= mg + 2Eq+\frac{Fel}{tg\alpha}+Fmag+mg+\frac{mv^2cos\alpha}{l} $$
$$ T= 2mg+2Eq+qvB+\frac{2Kq^2}{d^2tg\alpha}+\frac{mv^2cos\alpha}{l}$$
On the imminence of movement:
$$M\mu g=2T=4mg+4Eq+2qvB+\frac{4Kq^2}{d^2tg\alpha}+\frac{2mv^2cos\alpha}{l}$$
$$\mu=\frac{4mg+2q(2E+vB)}{Mg}+\frac{2}{tg\alpha Mg}(\frac{2Kq^2}{d^2}+\frac{mv^2sen\alpha}{l})$$
Where am I missing?