# Frictionless pulley- incline

by physics1234
Tags: frictionless, incline, pulley
 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.
 P: 482 This might be oversimplifying it a bit but shouldnt f=ma work?
 P: 21 What would you use as the mass? 98 isn't the right answer.
 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?
 P: 482 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)
 P: 15 Wouldn't they be moving in equilibrium? The string is not growing shrining, they are just shifting arent they?
 P: 21 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 definately the right answer, i just can't figure out the second part.
P: 1
 Quote by physics1234 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 definately 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

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