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

- 1018

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

- chestercat

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- 1018

##### 1 Attachment

- welshfella

Not sure about number 3
but i can help with number 4

- welshfella

|dw:1442848903125:dw|

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## More answers

- welshfella

let 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

solve these 2 equations to find T and a

- welshfella

you can find the speed of the bodies after 3 cm by using one of the equations for constant acceleration

- welshfella

you can use g = 9.81N

- 1018

i got 2.97 for acceleration

- 1018

is that correct? i got 2 equations from each block then combined those two equations

- welshfella

i got 0.794 for acceleration

- welshfella

and Tension = 11.72N

- 1018

can i see your work? thanks

- 1018

i have a question. in block B, i made the equation Tension - Weight = ma

- 1018

T=W+ma

- welshfella

T - 9.933 = 2.25a
12.753 - T = 1.3a
adding eliminates T
2.82 = 3.55a
a = 0.794 ms-2

- 1018

yes, but shouldnt be tension positive, and weight is negative? or am i wrong with my analysis?

- welshfella

you have to take account of the friction
the frictional force = 0.45 * wieght

- 1018

no, for block B. there's no friction

- welshfella

the tension is positive for 2.25 block and negative for block B

- welshfella

oh sorry block B force = weight - T

- 1018

im still confused. haha. wait let me draw

- 1018

|dw:1442851287028:dw|

- 1018

the direction of the tension is upward right, and weight is downward. so i thought its summation of forces would be T-W=ma

- 1018

or is this analysis wrong?

- welshfella

the resultant force acts downwards
weight acts down
tensions acts up

- 1018

so, tension is positive right, if it acts upward?

- welshfella

No it w - t = ma

- welshfella

don'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

haha, 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

- welshfella

yes

- 1018

and then for speed, just use one of the constant acceleration formulas?

- welshfella

yea m use v^2 = u^2 + 2as

- welshfella

u = iniital velocity = 0

- welshfella

sorry i gotta go right now
i think you can finish it OK

- 1018

thanks!

- welshfella

yw

- welshfella

For 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

wait let me check 3 again

- 1018

the first thing i did is to make resultant forces = 0 for it is constant speed. is that right?

- welshfella

yes thats what i meant by equilibrium

- 1018

oh i think i see what you're saying. my solution is not like that. it has frictions

- welshfella

well frictions must play a part

- 1018

yep but the F is not equal to the tension in my solution

- welshfella

the force must be just enough to overcome the frictions

- 1018

but the strings on A and B should be equal right? they have the same tension

- welshfella

yes

- welshfella

is the force equal to the sum of the frictional forces on A and B?

- welshfella

that would be 1.8*0.3 + 4.6*0.3?

- 1018

here's what i did : Block B : Fx = -F + Tension + friction 1 + friction 2 = 0

- 1018

pretty much yes, but i included tension

- welshfella

yea to be honest i'm not sure of this one.

- 1018

haha, that's ok i'll try to review with other questions first. haha. thanks though! big help.

- welshfella

yw

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