Page 86 - Schaum's Outline of Theory and Problems of Applied Physics
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CHAP. 6]                                 FRICTION                                      71



                  The frictional force to be overcome is
                                                                     ◦
                                 F f = µN = µw cos θ = (0.18)(5000 tons)(cos 10 ) = 886 tons
              The component of the ship’s weight parallel to the ways is
                                      F w = w sin θ = (5000 tons)(sin 10 ) = 868 tons
                                                               ◦
              Therefore an additional force of
                                        F f − F w = 886 tons − 868 tons = 18 tons
              is required.




                                     Multiple-Choice Questions


         6.1. Relative to the force needed to keep a box moving at constant velocity across a floor, to start the box moving usually
              needs
              (a) less force
              (b) the same force
              (c)  more force
              (d) any of the above, depending on the natures of the surfaces in contact

         6.2. When two surfaces are in contact, the frictional force between them depends on which one or more of the following?
              (a) the normal force pressing one surface against the other
              (b) the areas of the surfaces
              (c)  whether the surfaces are stationary or in relative motion
              (d) whether a lubricant is used or not
         6.3. A brick has the dimensions 8 cm × 16 cm × 32 cm. The force of starting friction between the brick and a wooden
              floor is
              (a) a maximum when the brick rests on the 8 cm × 16 cm face
              (b) a maximum when the brick rest on the 8 cm × 32 cm face
              (c)  a maximum when the brick rests on the 16 cm × 32 cm face
              (d) the same regardless of which face it rests on

         6.4. The coefficient of kinetic friction between two oiled steel surfaces is
              (a)  0.03   (c)  0.03 N/kg
              (b)  0.03 N  (d ) 0.03 kg/N

         6.5. A 60-N force is needed to start a 60-kg skater moving across a frozen lake. The coefficient of static friction for steel
              on ice is approximately
              (a)  0.06  (c)  0.6
              (b)  0.1  (d )1
         6.6. The coefficient of static friction for wood on concrete is 0.6. The force needed to set a 40-kg crate in motion on a
              concrete floor is
              (a)  2.4 N  (c)  235 N
              (b)  24 N  (d ) 653 N

         6.7. A 153-kg engine on wooden skids is resting on a level floor. The coefficients of static and kinetic friction are,
              respectively, 0.5 and 0.4. When two men push on the engine with a total horizontal force of 500 N, the frictional force
              that acts on the skids is
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