P16_017.PDF

17. The maximum force that can be exerted by the surface must be less than µ s N or else the block will not

follow the surface in its motion. Here, µ s is the coecient of static friction and N is the normal force

exerted by the surface on the block. Since the block does not accelerate vertically, we know that N = mg ,

where m is the mass of the block. If the block follows the table and moves in simple harmonic motion,

the magnitude of the maximum force exerted on it is given by F = ma m = mω 2 x m = m (2 πf ) 2 x m ,where

a m is the magnitude of the maximum acceleration, ω is the angular frequency, and f is the frequency.

The relationship ω =2 πf was used to obtain the last form. We substitute F = m (2 πf ) 2 x m and N = mg

into F<µ s N to obtain m (2 πf ) 2 x m <µ s mg . The largest amplitude for which the block does not slip is

x m =

(2 πf ) 2 = (0 . 50)(9 . 8m / s 2 )

=0 . 031m .

(2 π

2 . 0Hz) 2

A larger amplitude requires a larger force at the end points of the motion. The surface cannot supply

the larger force and the block slips.

µ s g