Consider the system described in the example with m1 = 0.51 kg and m2 = 0.40 kg.
ID: 1581836 • Letter: C
Question
Consider the system described in the example with m1 = 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 30.0°. 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.
F sin e a 5 | 7RF cos NNI *1* mg - m8 mg ba aExplanation / Answer
F min acts when friction and F will help to keep the system at rest
hence,
(0.51 kg x 9.81) = Fcos30° + 0.45(0.4 kg x9.81 - Fsin30°)
hence, Fmin = 5.05 N
for F max is when friction opposes F
Hence,
0.51 kg x 9.81 = Fcos30° - 0.45(0.4 kg x 9.81 - Fsin30°)
hence, Fmax = 6.20 N