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# Thread: Physics Help - Tension/Others (REPS)

1. ## Physics Help - Tension/Others (REPS)

Thanks for any help guys. Definitely will rep you a lot. Explanations would be nice too, I want to learn how to do these, especially the tension problems.

Problem 1)

The two blocks in the figure (Intro 1 figure) are connected by a heavy uniform rope with a mass of 4.00 kg . An upward force of 200 N is applied as shown. The top block has mass 6 kg, the bottom 5 kg.

A) What is the acceleration of the system in meters/seconds^2?

B) What is the tension at the top of the heavy rope in Newtons?

C) What is the tension at the midpoint of the rope in Newtons?

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Last Problem

The position of a 2.75 * 10^5 N training helicopter under test is given by
r = (0.02 m/s^3)*t^3 ihat + (2.2 m/s)*t jhat - (0.060 m/s^2)*t^2 khat

Find the net force on the helicopter at t =5.0 s.
Express your answer in terms of \hat i, \hat j, \hat k. Express your coefficient using two significant figures. In newtons.

2. strong

3. Anybody know either of these two problems?

4. Alright, well F=ma, right? I can't see the picture in the first one, so I can't really go anywhere from there, but I guess you have to find the total force applied to the rope and then use F=ma to find the acceleration. Though I suppose you probably already knew that.

I can't say I remember how to do the second one at all, I haven't done that stuff in three years. If I find my formula sheet from physics, I'll try to help you out.

Thanks for any help guys. Definitely will rep you a lot. Explanations would be nice too, I want to learn how to do these, especially the tension problems.

Problem 1)

The two blocks in the figure (Intro 1 figure) are connected by a heavy uniform rope with a mass of 4.00 kg . An upward force of 200 N is applied as shown. The top block has mass 6 kg, the bottom 5 kg.

A) What is the acceleration of the system in meters/seconds^2?

B) What is the tension at the top of the heavy rope in Newtons?

C) What is the tension at the midpoint of the rope in Newtons?

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I got a B in Phys last semester so Ill give it a try

A. Total mass of the whole system: 15kg
Total force by gravity applied on system: 15kg * 9.8 = 147N
Net force on the system: 200N - 147N = 53N going up
Force = acceleration x mass
Acceleration = Force / mass = 53N/15kg = 3.5 m/sec^2
B. Tension @ top is total weight of the rope and the block below = (4kg+5kg) x 9.8 = 88.2N
C. Tension @ midpoint is weight the block below and half of the rope= (5kg+2kg) x 9.8 = 68.6N

6. I think the acceleration is 3.5m/s^2 is that right?

7. Originally Posted by PunkGoesSwoll
I got a B in Phys last semester so Ill give it a try

A. Total mass of the whole system: 15kg
Total force by gravity applied on system: 15kg * 9.8 = 147N
Net force on the system: 200N - 147N = 53N going up
Force = acceleration x mass
Acceleration = Force / mass = 53N/15kg = 3.5 m/sec^2
B. Tension @ top is total weight of the rope and the block below = (4kg+5kg) x 9.8 = 88.2N
C. Tension @ midpoint is weight the block below and half of the rope= (5kg+2kg) x 9.8 = 68.6N
You got A right. B and C are wrong ( I already tried putting in 68.6, didn't work). I also tried 31.77 for B which didn't work. (4 + 5) * 3.53.

8. B. 31.5
C. 24.5

?

Edit: oh... i never got tension problems right... haha

9. try...

B. 56.7
C. 44.1

10. Originally Posted by PunkGoesSwoll
B. 31.5
C. 24.5

?

Edit: oh... i never got tension problems right... haha
Nope, not 24.5

11. try

b) 53N
c) 53N

edit: nvm, system is moving

12. B. 141.2n
C. 121.6

13. Originally Posted by PunkGoesSwoll
B. 141.2n
C. 121.6
nope

Thanks for any help guys. Definitely will rep you a lot. Explanations would be nice too, I want to learn how to do these, especially the tension problems.

Problem 1)

The two blocks in the figure (Intro 1 figure) are connected by a heavy uniform rope with a mass of 4.00 kg . An upward force of 200 N is applied as shown. The top block has mass 6 kg, the bottom 5 kg.

A) What is the acceleration of the system in meters/seconds^2?

B) What is the tension at the top of the heavy rope in Newtons?

C) What is the tension at the midpoint of the rope in Newtons?

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a) sum of forces =ma, so force up-force down=ma, force up is 200N, force down is gravity, which is 15lbs x 9.8, subtracting the two gives u about 50, set that = to ma, so 50=(15)a, a= about 3.33
b) forces are the 200 N up, and the gravitational force on the specified part =(9kg)(9.8)= about 90 subtracting the two gives you 110 N
c) same process as in b, except weight of rope is half, so 200-(7kg)(9.8)= about 130 N

am i close?

Last Problem

The position of a 2.75 * 10^5 N training helicopter under test is given by
r = (0.02 m/s^3)*t^3 ihat + (2.2 m/s)*t jhat - (0.060 m/s^2)*t^2 khat

Find the net force on the helicopter at t =5.0 s.
Express your answer in terms of \hat i, \hat j, \hat k. Express your coefficient using two significant figures. In newtons.
so take second derivative of the position equation to give u derivative. Then plug in the time given.

now convert weight of helicopter to kg's, then multiply the acceleration vector u got by the weight.....there u go

16. Originally Posted by amerrocks
a) sum of forces =ma, so force up-force down=ma, force up is 200N, force down is gravity, which is 15lbs x 9.8, subtracting the two gives u about 50, set that = to ma, so 50=(15)a, a= about 3.33
b) forces are the 200 N up, and the gravitational force on the specified part =(9kg)(9.8)= about 90 subtracting the two gives you 110 N
c) same process as in b, except weight of rope is half, so 200-(7kg)(9.8)= about 130 N

am i close?
B is 120, don't know how that is it (got it off some forum that had that part done)

For C i entered 131. That wasn't right.

17. Originally Posted by amerrocks
so take second derivative of the position equation to give u derivative. Then plug in the time given.

now convert weight of helicopter to kg's, then multiply the acceleration vector u got by the weight.....there u go
I did. I got the weight as 28061.2 kg ( 2.75 x 10^5 / 9.8)

Then the acceleration components = .12t ihat + 0 - .12

then i multiplied each by 28061.2. still not it.

B is 120, don't know how that is it (got it off some forum that had that part done)

For C i entered 131. That wasn't right.
oh, ok that makes sense....acceleration is 3.333 everywhere, so Force up on the part is 9(3.333) which is about 30, force down is still abt 90. when I pictured it, i saw the block being attached to one of those spring scales u hang stuff from, and i realized that when you pull the scale up, the weight actually increases

I did. I got the weight as 28061.2 kg ( 2.75 x 10^5 / 9.8)

Then the acceleration components = .12t ihat + 0 - .12

then i multiplied each by 28061.2. still not it.
did u put the 5 seconds in, so your vector would be <.6, 0, -.12> then multiply that by weight?

20. Originally Posted by amerrocks
did u put the 5 seconds in, so your vector would be <.6, 0, -.12> then multiply that by weight?
yeah, and got 16836.7 ihat, 0, 3367.34 khat.

It asked for two sig figs (Express your answer in terms of \hat i, \hat j, \hat k. Express your coefficient using two significant figures.) so I put

1.7 x 104 i hat, 0.0, 3.4 x 10^3 khat

21. Anyone? Please! haha I have to have this by tonight.

22. Just need the helicopter one now guys, thanks.