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

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Homework Help Overview

The discussion revolves around a physics problem involving two masses connected by a string over a frictionless pulley, with one mass on an incline. The original poster attempts to find the tension in the string after calculating the acceleration of the mass on the incline.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law (f=ma) and question the choice of mass used in calculations. There are attempts to clarify the relationship between the tension in the string for both masses and the conditions of motion.

Discussion Status

Participants are actively exploring different interpretations of the problem, particularly regarding the acceleration and tension calculations. Some guidance has been offered regarding the equations to use, but there is no explicit consensus on the correct approach to finding the tension.

Contextual Notes

There is a lack of agreement on whether the system is in equilibrium or not, which affects the interpretation of the forces acting on the masses. The original poster's calculations and assumptions are being scrutinized, and there is mention of verifying results through external sources.

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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