Sunday, October 23, 2011
HW6 #1
Does anyone know how to solve #1? I have already tried change in energy over time, P=Fv, etc...
HW6 #17
Hey I am having a problem getting #17 to work out on the homework with my numbers. When I tried solving it with other people's numbers the same way, I got the right answer, but mine arent working out. I have an 18 kg child on a playground swing moving with a speed of 4.7 m/s at the lowest point, and the swing is 1 meter long. I need to find the angle the string makes with the vertical. First I tried using conservation of energy with K=U, so .5mv^2=mgh. The m's cancel out and I am trying to solve for h, height above the lowest point. I end up with .5(4.7^2)=9.81h, and when I solve for h, I get 1.12, which doesn't work because it is higher than the length of the string. I also tried solving it by setting cos(theta)=(centripital acceleration)/g, and the acceleration is greater than g so it doesnt work. Please let me know if I am doing something wrong! Thanks!
Sunday, October 16, 2011
HW5 #18
Hey everyone, I just wanted to share a helpful tip for #18. Be careful when converting cubic kilometers to cubic meters. It's not a 1:1,000 ratio because the value is cubed.
Friday, October 14, 2011
HW5 #14
For what length of time should the power be calculated? Is this question asking for the power generated per second/hour/day?
Sunday, October 9, 2011
HW4 #5
Can somebody please help me with this question? I was able to answer the three preceding steps, but I seem to be stuck on this one. I attempted to calculate the horizontal components of T1 and T2 using trigonometry, but this yielded an incorrect solution. I then tried adding these values together, but this was wrong too. Please help!
HW4 #15
If you are having trouble with this problem but you managed to do the other force problems, you might want to take another look at your force diagrams (free body diagrams).
The smaller block hangs over the edge of a table (over a pulley), and is attached to a string. There is a tension T upward, a force due to gravity (Fg = m*g) downward, and the block's resulting acceleration is downward. You use Newton's 2nd law to get an expression for the tension: ∑F=Fg-T=ma so, rearranging and writing Fg=mg, T=mg-ma.
Now look at the larger block. It has a force of gravity downward, a normal force upward, and 2 ropes pulling to the right. That makes it T to the right and another T to the right, so there is 2T pulling to the right (look at the pulley: it makes the string pull twice). With "right" as the positive direction, Newton's 2nd law becomes ∑F=2T=ma. Plug in the expression you had above for T and solve for the acceleration. "The rest is just algebra."
Hope that helps. Good luck.
The smaller block hangs over the edge of a table (over a pulley), and is attached to a string. There is a tension T upward, a force due to gravity (Fg = m*g) downward, and the block's resulting acceleration is downward. You use Newton's 2nd law to get an expression for the tension: ∑F=Fg-T=ma so, rearranging and writing Fg=mg, T=mg-ma.
Now look at the larger block. It has a force of gravity downward, a normal force upward, and 2 ropes pulling to the right. That makes it T to the right and another T to the right, so there is 2T pulling to the right (look at the pulley: it makes the string pull twice). With "right" as the positive direction, Newton's 2nd law becomes ∑F=2T=ma. Plug in the expression you had above for T and solve for the acceleration. "The rest is just algebra."
Hope that helps. Good luck.
HW4 #12
#12 seems to be giving some trouble too. I'll put up this post, but if it doesn't help please ask for more specific suggestions.
Peter Pan is the mass that moves downward vertically. The problem says she starts from vo=0, and you are given the distance and the time, so you can use the x= equation to find her acceleration. That is also the acceleration of the counterweight, which is the block that accelerates upwards on the incline.
Next look at Peter Pan to find the tension in the rope: Fg acts downward, T (the tension in the rope) acts upward, and Peter Pan's acceleration is downward. If you choose downward to be the positive direction (because Peter Pan is accelerating downward) then when you use Newton's 2nd law you get ∑F=Fg-T=ma. (Note that if you had chosen up to be positive then your right hand side would be negative: =-ma.) Relace Fg with m*g to get mg-T=ma, or rearrange to get T=mg-ma. Note that the mass is Peter Pan's mass, which is known, so you can calculate the tension since you already figured out the acceleration.
Now draw a force diagram (free body diagram) for the counterweight on the incline. The force due to gravity is Fg and points downward. The normal force points up-and-to-the-left, perpendicular to the incline. The tension in the rope pulls up-and-to-the-right, along the incline.
Since the counterweight will be accelerating along the incline, make a chart with forces parallel to the incline (it will have acceleration "a"), perpendicular to the incline (will have no acceleration), and "?" (idk) which is in some other direction. The normal force is perpendicular to the incline. The tension in the rope is parallel to the incline. The force due to gravity, Fg, is some other direction, so belongs in the "?" column. Figure out the components of Fg in the parallel and perpendicular directions: Fgsinθ in the parallel direction and Fgcosθ in the perpendicular direction.
Now you can use Newton's 2nd law to look at forces in the direction parallel to the incline (use the direction up-and-to-the right which is the direction of the acceleration as the positive direction): ∑F=T-Fgsinθ=ma And this is NOT equal to zero since there IS an acceleration. Replace Fg with m*g to get T-mgsinθ=ma and solve for the mass, which is the mass of the counterweight which is what you are looking for. "The rest is just algebra."
Hope that helps. Good luck.
Peter Pan is the mass that moves downward vertically. The problem says she starts from vo=0, and you are given the distance and the time, so you can use the x= equation to find her acceleration. That is also the acceleration of the counterweight, which is the block that accelerates upwards on the incline.
Next look at Peter Pan to find the tension in the rope: Fg acts downward, T (the tension in the rope) acts upward, and Peter Pan's acceleration is downward. If you choose downward to be the positive direction (because Peter Pan is accelerating downward) then when you use Newton's 2nd law you get ∑F=Fg-T=ma. (Note that if you had chosen up to be positive then your right hand side would be negative: =-ma.) Relace Fg with m*g to get mg-T=ma, or rearrange to get T=mg-ma. Note that the mass is Peter Pan's mass, which is known, so you can calculate the tension since you already figured out the acceleration.
Now draw a force diagram (free body diagram) for the counterweight on the incline. The force due to gravity is Fg and points downward. The normal force points up-and-to-the-left, perpendicular to the incline. The tension in the rope pulls up-and-to-the-right, along the incline.
Since the counterweight will be accelerating along the incline, make a chart with forces parallel to the incline (it will have acceleration "a"), perpendicular to the incline (will have no acceleration), and "?" (idk) which is in some other direction. The normal force is perpendicular to the incline. The tension in the rope is parallel to the incline. The force due to gravity, Fg, is some other direction, so belongs in the "?" column. Figure out the components of Fg in the parallel and perpendicular directions: Fgsinθ in the parallel direction and Fgcosθ in the perpendicular direction.
Now you can use Newton's 2nd law to look at forces in the direction parallel to the incline (use the direction up-and-to-the right which is the direction of the acceleration as the positive direction): ∑F=T-Fgsinθ=ma And this is NOT equal to zero since there IS an acceleration. Replace Fg with m*g to get T-mgsinθ=ma and solve for the mass, which is the mass of the counterweight which is what you are looking for. "The rest is just algebra."
Hope that helps. Good luck.
HW#4 Q14
I don't know if anyone else is having trouble with this problem, but I'm having trouble figuring out number 14. If anyone could give some helpful advice that would be awesome.
HW4 #3
I am getting questions about how to do #3 and the other tension problems. Here is a start; feel free to add more details if you think there is something missing here, or post a more specific question if this doesn't help quite enough.
A block is suspended from a long string. Since the block hangs from the middle and the angle on the right is the same as the angle on the left, it is reasonable to say that the tension T1 in the string up-and-to-the-left is the same as the tension T2 up-and-to-the-right (there is no reason why one of them would be a bigger force than the other).
The block is hanging from a vertical (up-down) string. Let's call the tension in the vertical string T. If you draw a force diagram for the block itself, you have Fg downward and T upward. From Newton's 2nd law and using up as your positive direction, ∑F=T-Fg=ma=0 and it =0 because the block is not accelerating. So T-Fg=0, and T=Fg=m*g, and you have a numerical answer for the tension T in the vertical string.
Now draw a force diagram (free body diagram) for the knot where all 3 strings attach. You have T downward, T1 up-and-to-the-left, and T2 up-and-to-the-right. Make a chart with columns for forces in the x direction, the y direction, and the "?" (idk) direction. T1 and T2 are in the ? direction, so figure out the x- and y-components of them. In question 2 you figured out the angle from the ceiling, so the horizontal (x) components of T1 is T1cosθ and T2 is T2cosθ, and the vertical (y) components are T1sinθ and T2sinθ.
Now you can use Newton's 2nd law in the vertical direction. If you choose up to be the positive direction again then ∑F=T1sinθ+T2sinθ-Fg=ma=0, and again it is =0 because the knot isn't accelerating. Since T1 is equal to T2 you can call them both T1, and solve for T1. That is the tension in the string.
Hope that helps. Good luck.
A block is suspended from a long string. Since the block hangs from the middle and the angle on the right is the same as the angle on the left, it is reasonable to say that the tension T1 in the string up-and-to-the-left is the same as the tension T2 up-and-to-the-right (there is no reason why one of them would be a bigger force than the other).
The block is hanging from a vertical (up-down) string. Let's call the tension in the vertical string T. If you draw a force diagram for the block itself, you have Fg downward and T upward. From Newton's 2nd law and using up as your positive direction, ∑F=T-Fg=ma=0 and it =0 because the block is not accelerating. So T-Fg=0, and T=Fg=m*g, and you have a numerical answer for the tension T in the vertical string.
Now draw a force diagram (free body diagram) for the knot where all 3 strings attach. You have T downward, T1 up-and-to-the-left, and T2 up-and-to-the-right. Make a chart with columns for forces in the x direction, the y direction, and the "?" (idk) direction. T1 and T2 are in the ? direction, so figure out the x- and y-components of them. In question 2 you figured out the angle from the ceiling, so the horizontal (x) components of T1 is T1cosθ and T2 is T2cosθ, and the vertical (y) components are T1sinθ and T2sinθ.
Now you can use Newton's 2nd law in the vertical direction. If you choose up to be the positive direction again then ∑F=T1sinθ+T2sinθ-Fg=ma=0, and again it is =0 because the knot isn't accelerating. Since T1 is equal to T2 you can call them both T1, and solve for T1. That is the tension in the string.
Hope that helps. Good luck.
Subscribe to:
Posts (Atom)