Finding the acceleration of an object problem

  • Thread starter DJW
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  • #1
DJW
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Hey, I was needing some help with a friction problem involving finding the acceleration of an object.

A 50.0-kg box is being pushed along a horizontal surface by a force of 250 N directed 26 degrees below the horizontal. The coefficient of kinetic friction between the box and the surface is 0.300. What is the acceleration of the box?

I tried adding up all the forces in the X direction but I am still missing something. I know that the sum of the X forces is 44.8, and that the acceleration is .894 m/s², but I am not sure how the sum of the X forces is 44.8. Does the angle being below the horizontal have anything to do with it?

cos(26) = .899
sin(26) = .438
Fx = 250 N
Coefficient of Kinetic Friction = .300
 
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Answers and Replies

  • #2
hage567
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What about the normal force? You need that to figure out the frictional force which is acting in the -x direction. Sum up the forces in the y direction to find the normal force (hint it will not be just mg).
 
  • #3
DJW
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Would that be mgcosΘ?
 
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  • #4
hage567
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No. There are three forces acting in the y direction. Can you name them? What direction is each one acting in? Don't forget that the applied force is at an angle. That's important. You must resolve the applied force into its x and y components.
 
  • #5
DJW
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Hmm...250N, mg, and N? Not sure about the third one.
 
  • #6
hage567
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Almost. You will have weight, the normal force, and a component of the applied force.

Since the 250 N is applied at an angle, there will be a component of that force acting downwards in the y direction, and a component acting in the x direction. To find the normal force properly, you must find the component of that force that is acting in the y direction. You do this using trig. Do you know the trig relations of a right angled triangle? That's what you need. Draw a diagram, to see what goes where.

See: http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec5

Look at this page on friction, particularily the part on normal forces (near the bottom).
http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html#fri

They might help.
 
  • #7
DJW
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Oh, right. I already had the sum of the Y forces, but I guess I wasn't paying attention. I did mg + sin(26) x 250 = 599.59. Do I find ΣFx, add the two components, then solve with F=ma?
 
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  • #8
hage567
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Now that you have the normal force, you can find the frictional force. Do you know how to do that?
 
  • #9
DJW
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Now that you have the normal force, you can find the frictional force. Do you know how to do that?

Fk=μkN, right? (.30 x 599.59=179.877)?
 
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  • #10
hage567
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Yes, that's right. So do you know what to do next?
 
  • #11
DJW
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I subtracted that from 250 x cos(26) = 224.7 and got 44.8. Then I did F=ma which is 44.8=50(a) and got a=.896 (multiple choice answer is .894, which is the correct answer.). I had the normal force all messed up. Thanks for the help.
 
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  • #12
hage567
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Glad I could help. :smile:
 

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