Newton's Law and a rope

In summary, the system experiences an acceleration of 3.53m/s2 due to the force of 200N applied to the system. The tension at the top of the heavy rope is 120N, while the tension at the midpoint of the rope is 93.3N.
  • #1
Edwardo_Elric
101
0

Homework Statement


The two blocks are connected by a heavy uniform rope with a mass of 4.00kg. An upward force of 200N is applied as shown.
a.) What is the acceleration of the system?
b.) What is the tension at the top of the heavy rope?
c.) What is the tension at the midpoint of the rope?
The right is my own drawing of FBD and the sketch on the left side is from the book
Freebodydiagrams-1.jpg

Homework Equations


[tex] w = mg[/tex]
[tex] F = ma[/tex]

The Attempt at a Solution


b.)Acceleration of the system:
[tex] \sum{F_{net}} = F + (-W_{m1}) + (-W_{rope}) + (-W_{m2})[/tex]
Fnet = 200N - (6.00kg)(9.80m/s^2) - 4.00kg(9.80m/s^2) - (5.00kg)(9.80m/s^2)
Fnet = 53.0N
m_total = 4 + 5 + 6
m_total = 15kg
a = (Fnet)/(mtotal)
a = 53N / 15kg
a = 3.53m/s^2

b.) tension at the top of the heavy rope?
i don't know why the answer is 120N

c.) midpoint of rope,...
same here 93.3N
 
Last edited:
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  • #2
Your acceleration looks good for the system... but when you take your FBD's of the parts, you are forgetting to include the rope tension and/or rope mass. For sample, isolate the top block just at the top of the rope, and you have the 200N force up, the block weight down, AND the tension at the top of the cord down...
 
  • #3
so 200N is the upward force
and you need to find the total downward force of the 6.00kg block which is the tension of
the upper rope?

b.)
[tex] F_{y} = F_{applied} - W_{6kgblock} - F_{6kg block}[/tex] <<< is the force of 6 kg block negative?
F_{y} = 200 - (6.00kg)(9.8m/s^2) - (6.00kg)(3.53m/s^2)
= 120N

c.)
still confused in part C
i think i did this:
T = 5kg(3.53m/s^2) + 5kg(9.8m/s^2)
T = 66.65N

For computing the midpoint of the tension of the rope:
Average Force = (120N + 66.65N) / 2
= 93.3N
is the acceleration of the system negative?
 
Last edited:
  • #4
Are my answers correct? they agree at the back of my book but I am not sure if my solutions/assumptions here are correct..

thanks!...
 
  • #5
Last question

Befor i finalize my answer:
is my FBD correct?

FBD.jpg
 
  • #6
Edwardo_Elric said:
so 200N is the upward force
and you need to find the total downward force of the 6.00kg block which is the tension of
the upper rope?

b.)
[tex] F_{y} = F_{applied} - W_{6kgblock} - F_{6kg block}[/tex] <<< is the force of 6 kg block negative?
F_{y} = 200 - (6.00kg)(9.8m/s^2) - (6.00kg)(3.53m/s^2)
= 120N

c.)
still confused in part C
i think i did this:
T = 5kg(3.53m/s^2) + 5kg(9.8m/s^2)
T = 66.65N

For computing the midpoint of the tension of the rope:
Average Force = (120N + 66.65N) / 2
= 93.3N
is the acceleration of the system negative?
You're getting the right answers, but you're means of arriving at these answers is going to mess you up in the long run. And tension in ropes should not be considered as a 'force ' of the block.

Let's look at (b) first. To get the tension in the top of the rope, you draw an FBD of the top block by circling a 'cloud' around it. Wherever that 'cloud' cuts thru a contact force, there will be forces acting, which are in addition to the weight force on the block which is always there. So, in the FBD of the top block, you have 3 forces acting:

The applied force of 200N acting up (positive),
the weight of the block, 58.8N, acting down (negative)
the tension at the top of the rope, T_top, acting down (negative).
(Note that tension forces always pull away from the isolated object).
(Also note that I have chosen up as the positive direction consistent with the positive upward direction of the acceleration).

Thus, the NET force acting on the top block is 200 - 58.8 - T_top.

F_net = ma
200 -58.8 - T_top = (6)(3.53)
200 - 58.8 - T_top = 21.2
T_top = 120N.

Now in part (c), you have first calculated the tension in the bottom of the rope correctly, but please be consistent in your FBD. Isolating the bottom block, you have the tension at the bottom of the cord acting up (away from the object, positive), and the block weight, 49N, acting down (negative). Thus,
F_net = T_bot -49

F_net = ma
T_bot - 49 = 5(3.53)
T_bot = 66.6N

So then you took the average to find T_midpoint, which is OK in this problem due to the linearly varying tension, but in failing to recognize that, it is perhaps better instead to take a free body 'cloud' of the lower block that cuts through the mid-point of the rope, then apply Newton 2 again noting that you must include 1/2 of the rope's weight and mass:
F_net = ma
T_mid - 49 - 2(9.8) = (5 +2)(3.53)
T_mid = 93.3N
 
  • #7
thanks phantomjay

thank you so much
 

1. What is Newton's First Law of Motion?

Newton's First Law of Motion, also known as the Law of Inertia, states that an object at rest will remain at rest and an object in motion will remain in motion at a constant velocity, unless acted upon by an external force.

2. How does Newton's Second Law of Motion apply to a rope?

Newton's Second Law of Motion states that the force applied to an object is equal to the mass of the object multiplied by its acceleration. In the case of a rope, the force applied to it will determine how much it will stretch or move when pulled.

3. What is the relationship between force and tension in a rope?

In a rope, tension is the force that is transmitted through the rope when it is pulled on at both ends. The greater the force applied to the rope, the greater the tension will be.

4. Can Newton's Third Law of Motion be applied to a rope?

Yes, Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. In the case of a rope, when a force is applied to one end of the rope, the other end will experience an equal and opposite force.

5. How does friction affect the motion of a rope?

Friction is a force that opposes motion, so it can affect the movement of a rope. In the case of a rope being pulled, friction between the rope and the surface it is resting on can cause it to stretch less or move more slowly.

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