Hand on splint
Day 1 of camp.
Rockwall.
Can't wait for it to be over.
Day 2 of camp.
Rockwall.
Don't eat so much McDonald's.
I can't handle it anymore.
Day 3 after camp.
The strain.
Pain should be over pretty soon.
Strenuous activities are like that.
Day 7 after camp.
The strain.
Pain should be over pretty soon.
Strenuous activities are like that.
Day 14 after camp.
The strain straining.
Pain is not over.
"Better see a doctor."
Day 23 after camp.
Chronic sprain.
"See specialist."
Day 25 after camp.
Tendon inflammation.
Wait for swelling to subside.
"Rest or it will never get well."
Day 38 after camp.
Post-camp suffering.
Rest.
Thinking like an intellectual
A post on notsniw's Bloggie just rekindled my love for physics.
Ha, sprouted some nonsense in his pages based on my humble understanding of what I studied in my 4 years of university.
His questions as follows:
Q1: If a person exerts a force of 50N on a box and the friction involved is 4N, we know that the box will start moving because it has a resultant force of 46N acting on it. But... we know that from N3L, the box will exert an equal and opposite force which should cancel out with your initial 50N exerted, giving the box a resultant force of 0N. What is the problem here?My Answer:
Newton's 1st law states that "An object that is not moving will not move until a force acts upon it.
An object that is in motion will not change velocity until a force acts upon it."
Essentially this is also the law of inertia which indicates that the net force on an object is the vector sum of all the forces acting on the object. If this sum is zero, the state of motion of the object does not change. But in the case of Q1, the sum is not zero.
As a result of Newton's 1st law, we arrive at F = ma where it is defined in physics that the mass of a body is "reciprocally proportional to the bodies".
Derivation as follows:
F = k d(mv)/dt, where k = constant of proportionality, F and v in vector form
When mass is known to be constant,
F = kma, where F and a in vector form
Using SI units where k = 1 or unity,
F = ma, where F and a in vector form
Therefore F=ma (Newton's 2nd law) becomes a quantitative restatement of Newton's 1st law. Now mass is a measurement of inertia.
Then comes Newton's 3rd law. Even in the case of the 2 forces being equal in magnitude but opposite in direction, the accelerations are not. Based on Newton's 2nd law, the less massive object will have a greater acceleration. Example: Compare the force on the ground hit by the ball and the Earth's force on the ball. The magnitude is the same and opposite but due to Newton's 2nd law, the mass of the ball is smaller, thus a greater acceleration of the ball as compared to the Earth.Q2: To correctly identify the pair/pairs of action-reaction forces in a diagram involving a box on a table. Is the weight of the box an action-reaction pair with the contact force of the box on the table?My Answer:
Newton's 1st law: An object that is not moving will not move until a force acts upon it. No other force acting on either body. As mentioned by toukarin, I also interpret a closed system.
Newton's 2nd law:
box: F=ma, where a = constant
table: F=ma, where a = constant
Formula verfies what Newton's 1st law suggested, neither box nor table moves.
Newton's 3rd law: Constant acceleration in both bodies resulting in no motion imply mutual actions of 2 bodies upon each other. The changes made by these actions are equal, thus changes of the velocities made toward contrary parts are reciprocally proportional to the bodies, fulfilling F = k d(mv)/dt, in turn fulfilling F = d(mv)/dt = m(dv/dt) + v(dm/dt), conservation of momentum preserved.
Answer is yes.Never mind if my answers are correct or not. I like the thought processes. I have NOT felt intellectual for a really long time, coz involved in kiddy stuff for too long. I need more intellectual challenges.
Thanks, notsniw!
