2 Blocks 1 Pulley - First Time Poster

  • #1
Hi! First Time Poster.

Thank you for helping me with my problem!

Homework Statement



A 28.3-kg block (m1) is on a horizontal surface, connected to a 5.5-kg block (m2) by a massless string as shown in the Figure. The pulley is massless and frictionless. A force of 208.9 N acts on m1 at an angle of 29.7o. The coefficient of kinetic friction between m1 and the surface is 0.201. Determine the upward acceleration of m2.

\
\
\
M1-------O
|
|
M2

I'm not sure how to post pictures within my message just yet.

Homework Equations



X and Y components of the force on F1

Sum F = M * A

Kf = uK * Fn

g = 9.8m/s


The Attempt at a Solution



**Y+ is upwards and X+ is to the right**

I found the X and Y components of the Force on M1

I added M1 and M2... not sure if I need this but just in case.

And I drew Free Body Diagrams for both M1 and M2

- M1 has Normal Force acting Y+, which equals its Weight (m*g)... M1 also has Friction and Tension acting X+... the x component of the force acting X-

- M2 only has Tension acting Y+, and Weight (mg)... I suspect the tension to be equal to the tension... but I'm not sure.

Thanks! I hope I gave you enough information!
 

Attachments

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Answers and Replies

  • #2
I'm not really sure where to go from here. Any direction would be appreciated.

Thank you!

Also, can anyone see this post?? I'm not sure what the icon next to the post means.
 
  • #3
Can someone please message me? I've been awake and waiting for quite some time now...
 
  • #4
ehild
Homework Helper
15,543
1,914
Show your equations. The figure is very nice.

ehild
 
  • #5
Ok thank you for writing back!!! One second while i type it up!!!
 
  • #6
Known
m1 = 28.3
m2 = 5.5 kg
theta = 29.7
f = 208.9
uK = 0.201

Want


Acceleration of M2 in the Y direction...


________________

Fx = 208.9 cos (29.7) = 181.457
Fy = 208.9 sin (29.7) = 203.501

M1 + M2 = 28.3 + 5.5 = 33.8 (M3) (I'm not sure if I'll need this)



_____________________

For M2

Sum Fy = M*Ay

T - m*g = M* Ay


_____________________

For M1

F - T - Friction = M * A

F - T - uK * Fn = M * A




_________________

Thats all I have at the moment... :/
 
  • #8
ehild
Homework Helper
15,543
1,914
Use M1 and M2, respectively, in the equations for M1 and M2. They are not the same, and what is m in the equation for the hanging mass?
The length of the string does not change, so the magnitude of the vertical acceleration of M2 is the same as the horizontal acceleration of M1.
When you corrected the equations, add them: T will cancel and you can solve for the common acceleration a. Show it.

ehild
 
  • #9
M1: F-T-ukFn = m a

(208.9cos(29.7)) - T - 0.201*277.34 = 28.3*a

M2: T - mg = m a

T - 5.5*9.8 = 5.5 a

(T - (5.5*9.8) ) / 5.5 = a

___________________________________

How would I solve for something like this? Two variables?
 
  • #10
ehild
Homework Helper
15,543
1,914
You assumed that M1 moves on the left and M2 moves upward. It is not sure that true, but try. Then write:

M1:
F cosθ - T - μ Fn =M1a,

Fn=M1g-F sinθ, so

F cosθ - T-μ(M1g-F sinθ)=M1a.

M2:

T-M2g=M2a.

You can plug-in the numerical data at the end.

Add the equations in bold, or express T from the second one and substitute in the first one.

ehild
 
  • #11
208.9 cos 29.7 - (5.5 a + 28.3 * -9.8 ) - 0.201 ( 208.9 - sin (29.7) ) = 28.3 - a

a = 86.357????
 
  • #12
Yeah, this is not making sense... sorry. Can you please explain it again?
 
  • #13
Yeah, this is not making sense... sorry. Can you please explain it again?
 
  • #14
912
19
nerd, do as ehild said. "Add the equations in bold" it will cancel out T, then solve for a

EDIT: i can't emphasize the importance of a free body diagram enough. Always draw a FBD for different bodies in question.
believe me, life becomes easier when one does
 
  • #15
ehild
Homework Helper
15,543
1,914
208.9 cos 29.7 - (5.5 a + 28.3 * -9.8 ) - 0.201 ( 208.9 - sin (29.7) ) = 28.3 - a

a = 86.357????

Correct the red parts of the equation. g=9.8 m/s2, its direction was taken into account in the second equation. I can not imagine, why did you make 28.3-a from M2a. Isolate a, it is present at both sides. Take care of the parentheses.
It is better to solve a problem symbolically, and plug in data at the end.

ehild
 
Last edited:

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