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1018
 one year ago
please help with #s 3 and 4. thanks! (problems are in the picture below)
1018
 one year ago
please help with #s 3 and 4. thanks! (problems are in the picture below)

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welshfella
 one year ago
Best ResponseYou've already chosen the best response.1Not sure about number 3 but i can help with number 4

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442848903125:dw

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1let a be acceleration , t tension in the string, coefficient of friction = 0.45 equation of motion of block A_ Newtons 2nd law; T  0.45*2.25g = 2.25 * a equation of motion of block B: 1.3g  T = 1.3 * a

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1solve these 2 equations to find T and a

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1you can find the speed of the bodies after 3 cm by using one of the equations for constant acceleration

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1you can use g = 9.81N

1018
 one year ago
Best ResponseYou've already chosen the best response.0i got 2.97 for acceleration

1018
 one year ago
Best ResponseYou've already chosen the best response.0is that correct? i got 2 equations from each block then combined those two equations

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1i got 0.794 for acceleration

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1and Tension = 11.72N

1018
 one year ago
Best ResponseYou've already chosen the best response.0can i see your work? thanks

1018
 one year ago
Best ResponseYou've already chosen the best response.0i have a question. in block B, i made the equation Tension  Weight = ma

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1T  9.933 = 2.25a 12.753  T = 1.3a adding eliminates T 2.82 = 3.55a a = 0.794 ms2

1018
 one year ago
Best ResponseYou've already chosen the best response.0yes, but shouldnt be tension positive, and weight is negative? or am i wrong with my analysis?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1you have to take account of the friction the frictional force = 0.45 * wieght

1018
 one year ago
Best ResponseYou've already chosen the best response.0no, for block B. there's no friction

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1the tension is positive for 2.25 block and negative for block B

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1oh sorry block B force = weight  T

1018
 one year ago
Best ResponseYou've already chosen the best response.0im still confused. haha. wait let me draw

1018
 one year ago
Best ResponseYou've already chosen the best response.0the direction of the tension is upward right, and weight is downward. so i thought its summation of forces would be TW=ma

1018
 one year ago
Best ResponseYou've already chosen the best response.0or is this analysis wrong?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1the resultant force acts downwards weight acts down tensions acts up

1018
 one year ago
Best ResponseYou've already chosen the best response.0so, tension is positive right, if it acts upward?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1don't worry too much about positive or negative the point is that the motion is downwards and there for the net force acts downwards

1018
 one year ago
Best ResponseYou've already chosen the best response.0haha, ok. i'll just try it again. but when i get the acceleration, i can just sub it to one of the equations right? to get tension

1018
 one year ago
Best ResponseYou've already chosen the best response.0and then for speed, just use one of the constant acceleration formulas?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1yea m use v^2 = u^2 + 2as

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1u = iniital velocity = 0

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1sorry i gotta go right now i think you can finish it OK

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1For number 3 , as the speed is constant I think the force applied will equal the tension in the string. In other words the forces are in equilibrium. what do you think?

1018
 one year ago
Best ResponseYou've already chosen the best response.0wait let me check 3 again

1018
 one year ago
Best ResponseYou've already chosen the best response.0the first thing i did is to make resultant forces = 0 for it is constant speed. is that right?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1yes thats what i meant by equilibrium

1018
 one year ago
Best ResponseYou've already chosen the best response.0oh i think i see what you're saying. my solution is not like that. it has frictions

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1well frictions must play a part

1018
 one year ago
Best ResponseYou've already chosen the best response.0yep but the F is not equal to the tension in my solution

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1the force must be just enough to overcome the frictions

1018
 one year ago
Best ResponseYou've already chosen the best response.0but the strings on A and B should be equal right? they have the same tension

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1is the force equal to the sum of the frictional forces on A and B?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1that would be 1.8*0.3 + 4.6*0.3?

1018
 one year ago
Best ResponseYou've already chosen the best response.0here's what i did : Block B : Fx = F + Tension + friction 1 + friction 2 = 0

1018
 one year ago
Best ResponseYou've already chosen the best response.0pretty much yes, but i included tension

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1yea to be honest i'm not sure of this one.

1018
 one year ago
Best ResponseYou've already chosen the best response.0haha, that's ok i'll try to review with other questions first. haha. thanks though! big help.
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