Two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. The block in (Figure 1) has a mass m=10kg and is being pulled by a force F on a table with coefficient of static friction μs=0.3. Four forces act on it: The applied force F (directed θ=30∘ above the horizontal).
The force of gravity Fg=mg (directly down, where g=9.8m/s2).
The normal force N (directly up).
The force of static friction fs (directly left, opposing any potential motion).
If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as:

Fcosθ−μsN=0
Fsinθ+N−mg=0
In order to find the magnitude of force F, we have to solve a system of two equations with both F and the normal force N unknown. Use the methods we have learned to find an expression for F in terms of m, g, θ, and μs (no N).

Neutrons are placed in a magnetic field with magnitude 2.30 t. part a part complete what is the energy difference between the states with the nuclear spin angular momentum components parallel and antiparallel to the field? δe δ e = 2.77×10−7 ev previous answers correct part b part complete which state is lower in energy: the one with its spin component parallel to the field or the one with its spin component antiparallel to the field? which state is lower in energy: the one with its spin component parallel to the field or the one with its spin component antiparallel to the field? parallel antiparallel previous answers correct part c part complete how do your results compare with the energy states for a proton in the same field (δe=4.05×10−7ev)? how do your results compare with the energy states for a proton in the same field this result is smaller than but comparable to that found in the example for protons. this result is greater than but comparable to that found in the example for protons. previous answers correct part d the neutrons can make transitions from one of these states to the other by emitting or absorbing a photon with energy equal to the energy difference of the two states. find the frequency of such a photon. f f = mhz previous answersrequest answer incorrect; try again; 5 attempts remaining