Question
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?
Answers
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
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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