Friction No Clue

  • Thread starter randybrent
  • Start date
  • #1
Friction No Clue!!

A 2.5 kg block is pushed along a horizontal floor by a force of magnitude 14 N at an angle = -36° with the horizontal. The coefficient of kinetic friction between the block and the floor is 0.15. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block's acceleration.

Any help would be welcomed.
 

Answers and Replies

  • #2
486
1
C'mon, you have to have some clue!
What equations might be relevent?
What do you know?
What are you looking for?
 
  • #3
14n cos 36 - .15 =11.17 n ?

11.17/3.5= 3.19 m/s^2 ?
 
  • #4
486
1
First of all, no.
Second of all, I meant equations as in symbols.
like this:

v=x/t
x=123
t=.456

But yours is trickier. But not impossible:wink: .
 
  • #5
randybrent said:
A 2.5 kg block is pushed along a horizontal floor by a force of magnitude 14 N at an angle = -36° with the horizontal. The coefficient of kinetic friction between the block and the floor is 0.15. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block's acceleration.

Any help would be welcomed.

It helps to draw a FBD(free body diagram) of the block and write all the forces acting on it. I see 4 different forces, can you? (One of the force is at an angle, so that can be broken up into x and y components)
 
  • #6
w
i
l
l
uk -------------------- 14 cos 36
l
l
l
mg 14 sin 36

does this look right?
 
  • #7
486
1
Ummm... I don't get it.

About the equations...
you're looking for the force of friction.
The force of kinetic friction is given by the equaiton [tex]F_f=\mu_kN[/tex]
What do you know?
What are you looking for?
What do you still need to know?
Is there another equation you could use to find that?
 
  • #8
i dont know enough i guess
 
  • #9
Ff= .15*2.5kg*9.8= 3.675n ?
 
  • #10
486
1
OK, you're getting there.
Unfortunately, in this case N doesn't equal mg, even though it often does.
Use [tex] \Sigma F_y=ma=0[/tex], and notice that the force of the push has a y component.
 
  • #11
a= [.15*14sin36*9.8*2.5] /2.5 ?
 
  • #12
486
1
Look, Randy, can you do me a favor?
Don't just show me the numbers; state the equation symbolically first.

Also, I highly suggest that you solve for friction before looking for acceleration--you can't find acceleration without knowing the force of friction.
 
  • #13
i give up thank you for your time
 

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