Tension Problems Given Two Weights

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The discussion revolves around calculating the tension in cords supporting two hanging paint buckets, both weighing 3.2 kg. For the scenario where the buckets are at rest, the correct tension in each cord is approximately 31.6 N, but the initial calculations contained errors. When the buckets are accelerated upward at 1.6 m/s², the gravitational force must still be considered, leading to a revised approach for calculating the tensions. The key takeaway is that the gravitational force remains relevant even when the system is in motion, and understanding this principle clarifies the confusion. The participant ultimately gains clarity on how to approach the problem correctly.
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Homework Statement


One 3.2kg paint bucket(B) is hanging by a massless cord from another 3.2kg paint bucket(A), also hanging by a massless cord.
a.) if the buckets are at rest, what is the tension in each cord?
b.) if the two buckets are pulled upward with an acceleration of 1.6m/s2 by the upper cord, calculate the tension in each cord.

this is one of the problems I am confused with actually

Homework Equations


all i can make out is :
1. finding the tension when at rest
box[a] ==> FtensA = (agrav)(mA) + Ftens B
box ==> FtensB = (agrav)(mB)
2. finding the tension force given the acceleration 1.6m/s2
box[a]==>FtensA = mAaA +FtensB
box==>FtensB = mBaB

The Attempt at a Solution



I tried to solve for the tension when at rest and this is how it turned out:
FtensA = (9.8m/s2)(3.2kg) + FtensB
= 31.6N + 31.6N
= 63.2N
FtensB = (9.8m/s2)(3.2kg)
= 31.6N

Then I tried to solve for the tension given the acceleration
FtensA = (1.6m/s2)(3.2) + FtensB
= 5.12N + 5.12N
= 10.24N
FtensB = (1.6m/s2)(3.2kg)
= 10.24N

I really think that I'm wrong in here because its really confusing...Thanks In Advance! :)

<<Moderator note: Edited for formatting.>>
 
Last edited by a moderator:
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Part A you did correctly(though your calculation is a bit off). But when you take the tension of the cord given the new upward acceleration, remember that the first tension that you calculated doesn't go away. So how do you think you should approach part B with that in mind?
 
yandereni said:
FtensA = (agrav)(mA) + Ftens B

In this case, (agrav) is the gravitational acceleration which means that mA(agrav) is the gravitational force acting on the bucket A. Since the bucket is at rest, the equation is a force balance for bucket A (all forces must sum to zero).

yandereni said:
FtensA = mAaA +FtensB
The gravitational force does not disappear just because you are accelerating the bucket. Also, since the buckets are now being accelerated, the forces should not add to zero, but perhaps one of Newton's laws can tell you what it should sum to?
 
Oh, now it all makes sense. I forgot the part where the bucket is at rest. Thank you so much! I understand everything now. :)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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