Calculating Friction Force: Static and Kinetic Coefficients Explained

  • Thread starter wachaif
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In summary, the conversation discusses the necessary force to push a 6kg picture against a wall without it sliding, taking into account the static and kinetic friction coefficients between the wall and the picture. Different scenarios are considered, including pushing perpendicular to the wall and pushing at a 30 degree angle, both upward and downward. The necessary equations and steps are provided for each scenario.
  • #1
wachaif
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force question
i don't understand this..the way the forces go
if someone could show me how to do it and the steps..that would be great:
The static friction coefficient btw a wall and a pic is 0.7 while the kinetic one is 0.3. By what force shoudl one push the 6kg pic against the wall so that it will not slide?What if he/she pushed the pic with a force making an angle of 30 degrees to the horizontal (both cases)

1. Break the F into x-y component such as x=Fcos30 ,y = Fsin30
2.Since friction= U times normal force .and the normal force =Fcos 30
3.Set f=mg+Fsin30
U(Fcos30)=6(9.8) + F sin 30
0.7(Fcos30)=6(9.8) + F sin 30
F=553.58N

Am I correct ?
 
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  • #2
Well, I don't know. What was the question?

I put it that way because there are actually 3 different questions here.

1) What force is necessary if you push perpendicular to the wall?
Of course, there is a downward gravitational force of -mg= -6(9.8)= -58.8 Newtons and there has to be enough friction force to overcome that: friction force= UF= .7F. Set .7F= 58.8 and solve for F.

2) What force is necessary if you push at a 30 degree angle to the wall, angled UPWARD. the weight of the picture is still 58.8 Newtons.
The normal force will be F cos(30) so the friction force will be
0.7 F cos(30). Now there will be TWO forces vertically: the weight of 58.8 Newtons downward and the force F sin(30) upward of your push.
The net force downward is 58.8- F sin(30) and that must be offset by
0.7 F cos(30): Solve 0.7 F cos(30)= 58.8- F sin(30).

3) What force is necessary if you push at a 30 degree angle to the wall, angles DOWNWARD (the problem said "(both ways)"). THIS is the question you answered since you added F sin(30) to mg.
 
  • #3


Your calculations and steps are correct. To answer the first question, you would need to push the 6kg pic against the wall with a force of 553.58N to prevent it from sliding.

To answer the second question, where the force is applied at an angle of 30 degrees to the horizontal, you would need to break the force into its x and y components as you have done in step 1. Then, you would use the same equation as in step 3, but with the x and y components of the force (Fcos30 and Fsin30) instead of just F. This will give you the total force needed to prevent the pic from sliding at a 30 degree angle.

In general, to calculate the force needed to prevent an object from sliding, you would use the equation F=μN, where μ is the coefficient of static friction and N is the normal force (perpendicular force) acting on the object. Hope this helps!
 

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