Pully Problem with Friction

  • Thread starter ndwiseguy
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  • #1
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A 33.1- kg block (m1) is on a horizontal surface, connected to a 5.7- kg block (m2) by a massless string. The pulley is massless and frictionless. A force of 236.1 N acts on m1 at an angle of 30.9 °. The coefficient of kinetic friction between m1 and the surface is 0.217. Determine the upward acceleration of m2.

Now where I'm getting stuck is trying to determine the tension in the string between the two masses, because in either equation you have two unknowns, the tensile force and the acceleration. I thought that possibly merging the two equations could help. Am I on the right track, or completely off??
 

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  • #2
ndwiseguy said:
A 33.1- kg block (m1) is on a horizontal surface, connected to a 5.7- kg block (m2) by a massless string. The pulley is massless and frictionless. A force of 236.1 N acts on m1 at an angle of 30.9 °. The coefficient of kinetic friction between m1 and the surface is 0.217. Determine the upward acceleration of m2.

Now where I'm getting stuck is trying to determine the tension in the string between the two masses, because in either equation you have two unknowns, the tensile force and the acceleration. I thought that possibly merging the two equations could help. Am I on the right track, or completely off??
If you have two unknowns, you need at least two independent equations in order to solve for both. When you have your two equations, simply solve for one variable in terms of the other, and substitute that expression into the other equation so that it will become an equation in one unknown (substituting your expression back into the equation it was derived from will just yield a universal truth like 1=1 :smile: ). If you know a bit of linear algebra, you can also use a convenient linear combination of your equations to isolate a variable.
 
  • #3
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so my two equations are:
Fpcos(x)-Ff-T=m1a
T-m2g=m2a

Now i solved eqn two for T and got T=m2a+m2g which i then inserted into the first question. Now, for the force of friction, would i use 0.217*9.8*33.1 because the normal force would be equal to the force of gravity? If so, my answer is 2.786 m/s^2. Sound right?
 
  • #4
ndwiseguy said:
Now, for the force of friction, would i use 0.217*9.8*33.1 because the normal force would be equal to the force of gravity?
Not in this case. Draw a free-body diagram of m1 alone. There is more than one force normal to the friction surface.
 
  • #5
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The only other force I can see acting normal to the friction surface is the y-component of the pulling force. So would I use both that and mg??
 
  • #6
ndwiseguy said:
The only other force I can see acting normal to the friction surface is the y-component of the pulling force. So would I use both that and mg??
Yep. Remember to sign the components of your forces correctly. The normal force in the friction formula is oriented into the friction surface, so the y-component of your force detracts from it. :smile:
 
  • #7
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Fpcos(x)-(0.217*(Fpsin(x)-m1g))-(m2g+m2a)=m1a

Look right?? The final answer i got was 1.82 m/s^2
 
  • #8
ndwiseguy said:
Fpcos(x)-(0.217*(Fpsin(x)-m1g))-(m2g+m2a)=m1a

Look right?? The final answer i got was 1.82 m/s^2
The signs of the forces contributing to the normal force on the block are incorrect. We want greater friction to result with greater force being directed directly to the surface, so we want frictional force to be represented by -mu*(m1g - Fsin(x)).
 
  • #9
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perfect thank you very much
 

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