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Frictionless pulley incline 
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#1
Sep2506, 07:44 PM

P: 21

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. 


#2
Sep2506, 07:47 PM

P: 481

This might be oversimplifying it a bit but shouldnt f=ma work?



#3
Sep2506, 08:09 PM

P: 21

What would you use as the mass? 98 isn't the right answer.



#4
Sep2506, 08:46 PM

P: 15

Frictionless pulley incline
Well, I'm just wondering on this one...but isnt 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? 


#5
Sep2506, 09:04 PM

P: 481

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) 


#6
Sep2506, 09:12 PM

P: 15

Wouldn't they be moving in equilibrium? The string is not growing shrining, they are just shifting arent they?



#7
Sep2606, 02:07 PM

P: 21

I got the acceleration by a= (m2m1sinTHETA/m1+m2)g so it was (107sin43/17)(9.8) = 3.0126
We can check to see if we got them right on the internet so that's definately the right answer, i just can't figure out the second part. 


#8
Sep2706, 11:02 PM

P: 1

try using the equation T= m1(ga) = (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 


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