Posted by **Stan** on Wednesday, March 3, 2010 at 1:26am.

Figure 6-24 shows an initially stationary block of mass m on a floor. A force of magnitude 0.540mg is then applied at upward angle 16 degrees. What is the magnitude of the acceleration of the block across the floor if (a)Us=.6 and Uk=.5 and (b) Us=.4 and Uk=.3?

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**drwls**, Wednesday, March 3, 2010 at 7:04am
Is the 0.540mg 0.540 milligrams or 0.54m * g? I will assume the latter.

For case (a), the maximum possible static friction force is

Fs = (m*g cos16* - 0.54m*g sin16)*Us

= 0.487 m g

The horizontal component of applied force is 0.540m*g*cos 16 = 0.519 m*g. Since this exceeds the maximum possible static friction force, the block slips and accelerates at a rate determined by

kinetic friction.

The friction force in the horizontal direction must be recomputed and subtracted from the applied force component in that direction.

For case (b), proceed similarly

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