2007-10-30
Today's nonsense
- Homework due today and tomorrow, and calculus homework.
- RSS feed for the calculus/physics websites?
ex 3) A picture weighing 8 N is supported by two wires at angles of 60 degrees and 30 degrees to the ceiling. Find the tension in each wire.
Let's look at what's going on- you have to know alternate interior angles where if a line is cut by .... So now we have to do our balances of forces. We're still using right as positive and left as negative. So the 30 degree component will be positive and the other one will be negative.
Sum of F_x = T_2cos(30) - T_1cos(60) ---- this second part is negative because it's in the opposite direction of positive x direction.
Sum of F_y = T_2sin(30) + T_1sin(60) - mg ... and so since the picture is not moving both of these summed forces are going to be nothing.
So therefore the F=ma and a=0. Find the tension in each wire. You have two equations and two unknowns. Everything else you know, you know m, you know g, so we can take one equation and solve for, or we can try to do simultaneously, couldn't we just do where we put them next to each other and subtract? But it would be difficult. So let's take the first equation and solve for, what, T_1 or T_2? T_2 first.
So what is T_1? T_1 = ( T_2cos(30) ) / cos(60) ... so now we will substitute this in to the other equation and solve for T_2.
T_2sin(30) + T_2cos(30)sin(60)/sin(60) = mg.
But now what? Oh my. So we can plug in values right now. The only benefit to not plugging in values is getting you used to the degrees because in some problems you will be asked to do a totally general equation.
4) So we have a box. A 32 kg box hangs motionless from a vertical rope. It is then accelerated upward to a speed of 6 m/sec in 0.8 seconds. What was the tension in the rope?
Sum of F_x = 0.
Sum of F_y = T - mg = ma_y
What is the acceleration in the y direction? Change in velocity / change in time. So that's ( 6 m/sec ) / 0.8 sec = 7.5 m/sec^2. So this type of problem is one of those where, say, an AP test question free-response where it's part of the free-response, you could lose credit by not showing your work.
T = ma_y + mg
Remember, you are just looking at the forces, and that's the weight and the tension. So now you plug and chug. 553.6 N.
5) This one's pretty cool. Yeah. You could do this one easily. Next time you fly on a plane. All you need is a protactor with a string hanging down, and you just have to see as the airplane accelerates that the string will not hang straight down, so you can calculate the acceleration ...... As your plane speeds down the runway, you notice your suspended yo-yo's string makes a 22 degree angle with the vertical.
A) What is the acceleration of the plane?
B) If the mass of the yo-yo is 40 g, what is the tension in the string?
Sum of F_x = T_1sin(22) = ma_x
Sum of F_y = T_1cos(22) - mg = 0
So now we can figure out T_1 = mg / cos(22) = 0.423 N.
a_x = T_1sin(22) / m = 3.963 m/sec^2.
Nick pointed out this cool physics-problem helper relevant to this set of notes.