What is the tension in the string for a frictionless pulley-incline system?

AI Thread Summary
In a frictionless pulley-incline system with two crates, the acceleration of the 7.00 kg crate up the incline was calculated to be 3.0126 m/s². The tension in the string was initially difficult to determine, with various formulas attempted but yielding incorrect results. The correct approach for calculating tension involves using the equation T = m1(g - a), which led to a tension of 67.9 N for the 10.0 kg crate. The discussion highlighted the importance of understanding the dynamics of the system, particularly that the crates are not in equilibrium due to their movement. Overall, the focus was on accurately determining both acceleration and tension in the system.
physics1234
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Two packing crates of masses m1 = 10.0 kg and m2 = 7.00 kg are connected by a light string that passes over a frictionless pulley. The 7.00 kg crate lies on a smooth incline of angle 43.0°. Find the acceleration of the 7.00 kg crate up the incline. Find the tension in the string.

I found the acceleration (correctly) to be 3.0126 but I can't seem to get the tension part of the question. I tried the formula T=(m2)(g)(sinTHETA)+(m2)(a) using 7(3.0126)sin43+7(3.0126) and 10(3.0126)sin43+10(3.0126) but neither gave the correct answer.
 
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This might be oversimplifying it a bit but shouldn't f=ma work?
 
What would you use as the mass? 98 isn't the right answer.
 
Well, I'm just wondering on this one...but isn't the T of the string that is holding the block on the incline equal to the T of the string holding the hanging block?

So how would you figure out T on the hanging block?
 
well no because theyre moving theyre not in equilibrium, i assume?

just write out how you got the accel so we can see how ur goin about this as i agree with ur workin for T so maybe accel is wrong? ( i know you said you got it right but just check)
 
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Wouldn't they be moving in equilibrium? The string is not growing shrining, they are just shifting arent they?
 
I got the acceleration by a= (m2-m1sinTHETA/m1+m2)g so it was (10-7sin43/17)(9.8) = 3.0126

We can check to see if we got them right on the internet so that's definitely the right answer, i just can't figure out the second part.
 
physics1234 said:
I got the acceleration by a= (m2-m1sinTHETA/m1+m2)g so it was (10-7sin43/17)(9.8) = 3.0126

We can check to see if we got them right on the internet so that's definitely the right answer, i just can't figure out the second part.


try using the equation T= m1(g-a) = (10.0kg)[(9.80 - 3.0126) m/s^2] = 67.9N

i'm sure that's right cause i just did the same problem for hw.
good luck :cool:
 
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