SOLUTION: The coefficient of static friction between a 20-kg block and a horizontal surface is 0.25 and the coefficient of kinetic friction is 0.15. A cord is attached to the block, the cord

Algebra ->  Test -> SOLUTION: The coefficient of static friction between a 20-kg block and a horizontal surface is 0.25 and the coefficient of kinetic friction is 0.15. A cord is attached to the block, the cord      Log On


   



Question 1182784: The coefficient of static friction between a 20-kg block and a horizontal surface is 0.25 and the coefficient of kinetic friction is 0.15. A cord is attached to the block, the cord making an angle of 10 degrees with the horizontal. (a) What tension in the cord will just start the block? (b) What is the tension in the cord when the block moves with uniform velocity?
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
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The coefficient of static friction between a 20-kg block and a horizontal surface is 0.25
and the coefficient of kinetic friction is 0.15. A cord is attached to the block,
the cord making an angle of 10 degrees with the horizontal.
(a) What tension in the cord will just start the block?
(b) What is the tension in the cord when the block moves with uniform velocity?
~~~~~~~~~~~~~~~


The plot to the problem is shown in the Figure below.



    

    Figure.  The body on the horizontal floor



Decompose the given force  F  into horizontal  Fx  and vertical  Fy  components. 
Horizontal component  Fx  is equal to  F*cos(10°);  vertical component  Fy  is  F*sin(10°).


The reaction force  N  acts from the floor vertically up to the box.  Together with 

vertical component  Fy  it balances the weight:  W = Fy + N.   So,  N = W - Fy = mg - Fy. 



The condition of start moving is that the horizontal force component  Fx  is equal 
to the reaction force multiplied by the  static friction coefficient


    Fx = k%5Bstat_frict%5D%2AN,   or   F*cos(10°) = 0.25*(mg - F*sin(10°)).


From the last equation,  F*cos(10°) + 0.25*F*sin(10°) = 0.25*mg,  which gives for the magnitude of F


    F = %280.25mg%29%2F%28cos%2810%5Eo%29+%2B+0.25%2Asin%2810%5Eo%29%29 = %280.25%2A20%2A10%29%2F%280.985+%2B+0.25%2A0.174%29 = 48.614 N.    


So, the tension of the cord, the force F, to start moving is  48.614 newtons.

It is the answer to question (a).



The condition of moving with the constant velocity is that the horizontal force component  Fx  is equal 
to the reaction force multiplied by the  kinetic friction coefficient


    Fx = k%5Bkin_frict%5D%2AN,   or   F*cos(10°) = 0.15*(mg - F*sin(10°)).


From the last equation,  F*cos(10°) + 0.15*F*sin(10°) = 0.15*mg,  which gives for the magnitude of F


    F = %280.15mg%29%2F%28cos%2810%5Eo%29+%2B+0.15%2Asin%2810%5Eo%29%29 = %280.15%2A20%2A10%29%2F%280.985+%2B+0.15%2A0.174%29 = 29.67 N.    


So, the tension of the cord, the force F, to move the block uniformly is  29.67 newtons.

It is the answer to question (b).

Solved.

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In this solution,  I used the approximate value of  10 m/s^2  for the gravity acceleration
instead of the more precise value of  9.81 m/s^2.