To be sure understand it:
The locations of the masses on the rod below don't matter. All the total mass in all of the rods below (really all the weight) is in the string at the end of the current rod.
For rod b, you have 149*2 kg at x=0 cm. And 149 kg at x=65 cm. Divide by total mass (149 kg...
Oooh. Put that answer in part C to make sure my reasoning is correct. If it is, then part b should be:
(0cm * 2 * 149 kg + 65 cm * 149 kg * 1) / 3 * 149 kg.
No, it should be the mass*distance at the one end + mass times distance at the other end, divided by total mass.
Or [149 kg * (3) + 149 kg *(1) * 65 cm] / 4 * 149kg
the diagram shows that L3 is the distance from the end of that rod to the balance point, so you don't need to add anything.
If...
Eep, no. m as in m=mass=149 kg.
Rod 3 has 3*149 kg at one end, and 149 kg at the other. Because all of rod 2 and rod 1's mass (a total of 3*149 kg) is supported by one string at the end of rod 3.
The pulley is attached to m1. So they have to have the same acceleration when the rope between them is under tension. If the pulley moves 1 meter, then 1 meter of the m1 rope has moved on both sides of the pulley, so 2 meter of rope has gone past the pulley, so m2 moves twice as much as m1.
So when you say x3 you are saying "the number you get when x is multiplied by itself 3 times". You got that. So when you say s1/3, you are saying "the number you multiply by itself 3 times to get x".
Think about 27. 273 = 19683. 271/3 = 3. There are 9 "1/3's" in 3, right? 1/3 * 9 = 9/3 = 3...
Okay, now we can ignore 9.81 because it will be in every term.
But my point is, you can model rod 2 as having 1m (149 kg) at one end and 2m (298 kg) at the other.
So find the center of mass of that system for Rod 2.
Rod 3 can be modeled in a similar way, but you now have 3m on one end.
Well, that's torque. I just want to know if you held the string with rod 1 and its masses attached, instead of attaching it to the rod, how much force would be pulling down on your hand (ignoring the weight of your own hand)?
You got the right result, but just be careful. You divide G from both sides, not subtract. G is being multiplied by something, so you can't just subtract it out. But like i said, you got the right result anyway in this case, just be careful in the future.
We don't own your text book.
You'll have to post the full problem, preferably with a drawing, picture or scan of the diagram included.
You will also have to tell us how you've approached the problem so far.
I would start by solving everything that is in series with no junctions in between. That should clean it up a bit. If you still have trouble, post your drawing again after doing that.
You entered this into the calculator wrong. Watch your parenthesis.
A big clue that something is wrong is that you have a period of 2000 s. If light has a really small wavelength and a fast speed, does it make sense that it should take over half an hour for one wave cycle?
Keep in mind that x and y are separate.
So you know that in time T the ball must travel 110 feet in the x direction and 90 feet in the y direction.
Your equations are close, but v_i in the first equation should only be the y component of v_i.
the second equation is for the x direction, so...
Homework Statement
I just want to confirm this.
A block m1=10 kg is hanging from a rope. the rope is attached by a frictionless pulley to a block m2=5 kg on a frictionless slope with a 40 degree angle from the ground.
Find the tension in the rope and the acceleration of the first block...