Consider the system described in the example with m| 0.51 kg and m2-0.40 kg. The

Question

Consider the system described in the example with m| 0.51 kg and m2-0.40 kg. The coefficient of static friction between the block and the surface is 0.45. The angle of the force F is equal to 28.2. If F= 0, you can easily show that the block will accelerate to the left since the maximum static friction force is not sufficient to keep the block at rest. If F is sufficiently large, it is clear that the block will accelerate to the right. Find the range of F that allows the system to remain at rest. Fmin= 2.57 max= 5.67

Explanation / Answer

to keep from moving left,

Forces acting on m1 = 0 or

m2*g = *m1*g + Fcos28

F = g*(m2-m1)/cos28.2

= 9.8(0.40 -0.45*0. 51)/cos28.2 = 1.9 N(to keep it from moving to left)

Similarly to keep it from moving to right,

m2*g+m1*g = Fcos28.2 solve for F,

F = 9.8(0.40+0.45*0.51)/cos28.2 = 7 N

answer:

Fmin = 1.9 N

Fmax = 7 N

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