Question

#8-#9

Four identical 100 N boxes are initially at rest on a rough horizontal surface. In all cases, the coefficient of static friction between the box and the surface is 0.70 and the coefficient of kinetic friction is 0.25. Forces are applied as shown. 8. Circle the letter of the box (or boxes) subject to a frictional force of 25 N. A.C. P U 9. Circle the box(es) in equilibrium. A+B .70 25 N 50 N 12 Sa 100 N 100 N Esmax M, N (70) (100) : 70 Fx M x Ñ (25)400) = 25 с 75 N 125 N 25N 100 N 259100 N 10. A 70 kg skater moves on smooth ice. With how much horizontal force must she push on the ice to accelerate at 4.3 m/s?? 70 kg *4.3 mis2 : 30 1 N 11 After celerating how much horizontal force is required to glide with a constant horizontal

Answer 1

Solution -

Static friction is equal to

0.7*100 = 70 N (Maximum value )

Kinetic friction is equal to

0.25*100 = 25 N

In case A and B the applied force is less than the static friction's maximum value.

The static friction has the property of self adjustment upto maximum value till the applied force is less than the maximum value of static friction so in first case the A the static friction will be equal to 25 N and in second case B it will be 50 N . and hence these two boxes will not move and will remain in equilibrium.

In case C and D the applied force is more than the value of static friction(70 N - Maximum value ) and hence will begin to move and then Kinetic friction will come into play and hence the friction force applies here will be 25 N (Kinetic friction ) and these two boxes will move with the net force of 50 N and 100 N respectively.

Now comes to answer of your questions -

Q8. Solution - Box A , C , D will be subjected to the frictional force of 25 N.

Q.9 Solution - Box A and B will remain in equilibrium because the value of applied force is less than static friction hence they won't move .

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