Find tension between blocks 2 & 3

  • Thread starter Thread starter Rey4312
  • Start date Start date
  • Tags Tags
    Blocks Tension
AI Thread Summary
The discussion revolves around calculating the tension between three connected blocks with given masses, assuming no friction. The user initially sets up free body diagrams (FBD) for the blocks but struggles to determine the tension forces accurately. The community suggests labeling the tensions as "T1" and "T2" and emphasizes that the acceleration is the same for all blocks. The user realizes the importance of including the weight (W = mg) of the hanging block in their calculations. Ultimately, they find a way to incorporate the masses and acceleration into their equations to solve for tension.
Rey4312
Messages
13
Reaction score
0

Homework Statement



There are three blocks connected by strings as shown in the figure. Each block has the mass as following mass,
m1=0.8kg
m2=2.2kg
m3=1.7kg
Assuming there is no friction,


Homework Equations



F=MA

The Attempt at a Solution



In my FBD of the hanging box
\uparrow <--- Tension
[]
\downarrow <--- W=mg

^^ in y direction: T-W=0
T=W

The FBD of the box on the table
\uparrow <---Normal Force
[] \rightarrow <--- Tension
\downarrow <--- W=mg

^^ x direction: t=0
y direction: Fn-W=O

But I am completely confused on how to find the tension of the boxes
 

Attachments

  • Screen shot 2012-02-29 at 4.23.01 PM.png
    Screen shot 2012-02-29 at 4.23.01 PM.png
    17.7 KB · Views: 700
Physics news on Phys.org
Welcome to PF!

Hi Rey4312! Welcome to PF! :smile:

Call the tensions "T1" and "T2", and the acceleration "a" (it'll be the same for all three boxes, won't it? :wink:)

Then do F = ma three times, once for each box, and solve …

what do you get? :smile:
 
But if i do that the acceleration = 0 and so it makes all the forces 0... right??or am i missing something and it should be gravity?
 
Yes, you need to include W (=mg) for the hanging box.
 
W for the hanging box is 16.66

and W for box 2(m2) is 21.56...

Where do i go from there, because that doesn't give me acceleration...
 
i figured it out... i ended up having to do m2a+m1a
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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}...
Back
Top