Q.100. Statics, Inclined Plane, friction.

Question 100.

The coefficient of static friction between the 3.00 kg crate and the 35.0° angle incline is 0.300.

What minimum force F must be applied to the crate perpendicular to the incline to prevent it from sliding down the incline?

Answer 100.
If R = normal reaction from the incline on the crate, then
F + R = mgcos35°
=> R = (mgcos35° - F)

Maximum static frictional forc
= μ (mgcos35° - F)
This should be more than the downward component of weight, mg, i.e., mgsin35°
=> μ (mgcos35° - F) > mgsin35°
=> F > mg [(1/μ)sin35° - cos35°]
=> F > (3)(9.8)[(1/0.3)sin35° - cos35°]
=> F > 32.13 N.
=> Minimum force to be applied = 32.13 N.

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