Solve Friction Problem: Max Force on 1.0 kg Block

  • Thread starter Fusilli_Jerry89
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In summary, the maximum force that can be applied horizontally to the 1.0 kg block without letting the 0.5 kg block slip is 2.94 N. Both blocks have to be accelerating at the same rate, therefore the normal force acting on the bottom of the 1.0 kg block is 2.94 N. The equation for the first block is F - 3.675 = 1.0a and for the small block is 1.715 = 0.5a. After solving for acceleration, the final answer is 7.105 N.
  • #1
Fusilli_Jerry89
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A 0.5 kg mass is on top of a 1.0 kg mass on a table. The static coefficient of friction between the blocks is 0.35, and the kinetic coefficient of friction between the table and1.0 kg block is 0.20. What's the maximum force that can be applied horizontally to the 1.0 kg block without letting the 0.5 kg block slip?

the acceleration of the small block has to be zero, so the friction between the two blocks cannot exceed 1.715 N.I got the equation for the first blockdown to F-3.675=1.0a but what do I do after that? Does the acceleration of the large block have to be zero as well, because if it does then the answer is 3.675 N.
 
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  • #2
Fusilli_Jerry89 said:
the acceleration of the small block has to be zero, . . . .
No it does not. Think about the motion that will occur if there is no slipping.
 
  • #3
well won'tthe accelration of the small block havet to match that of the big one? Wouldn't that mean it is 0 in relation to the big block?
 
  • #4
So is the answer 7.105 Newtons then?
 
  • #5
Fusilli_Jerry89 said:
well won'tthe accelration of the small block havet to match that of the big one? Wouldn't that mean it is 0 in relation to the big block?
Yes they have to be the same. Yes that means it is zero relative to the big block, but the big block is accelerating. You don't know how to do problems in accelerating reference frames. Do the problem in the laboratory frame. Both blocks will be accelerating.
 
  • #6
Fusilli_Jerry89 said:
So is the answer 7.105 Newtons then?
I don't think so. How did you get it?
 
  • #7
well i found theat the equation for the large block was:
F-Ff-Ff=ma
F-1.96-1.715=1.0a
F-3.675=a

and for the small block:
Ff=ma
1.715=0.5a
a=3.43

then I substituted a into the first equation because they are accelerating at the same rate
F-3.675=3.43
F=7.105 N
 
  • #8
Why 1.96? What is the normal force acting on the bottom of the 1kg block?
 
  • #9
woops i forgot to add the 0.5 kg. So it would be 2.94 N instead?
 
  • #10
Fusilli_Jerry89 said:
woops i forgot to add the 0.5 kg. So it would be 2.94 N instead?
That looks better.
 

FAQ: Solve Friction Problem: Max Force on 1.0 kg Block

What is friction?

Friction is a force that resists the motion of an object when it comes into contact with another object or surface.

How does friction affect the motion of an object?

Friction can cause an object to slow down or come to a stop, or it can change the direction of motion. It can also cause wear and tear on surfaces that are in contact with each other.

How do you calculate the maximum force of friction on a block?

The maximum force of friction on a block can be calculated using the coefficient of friction and the normal force. The equation is: max friction force = coefficient of friction x normal force.

What factors affect the maximum force of friction?

The maximum force of friction is affected by the type of surfaces in contact, the roughness of the surfaces, and the normal force applied to the object.

How can you reduce friction and increase the maximum force on an object?

Friction can be reduced by using lubricants or by using smoother surfaces. The maximum force on an object can be increased by increasing the normal force applied to the object.

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