Asked by Cameron
www.webassign.net/sercp9/4-p-047-alt.gifThe coefficient of static friction between the m = 3.60−kg
crate and the 35.0° incline of the figure below is 0.330. What minimum force F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
crate and the 35.0° incline of the figure below is 0.330. What minimum force F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
Answers
Answered by
Damon
normal force = 3.6 * 9.81 * cos 35 + F
friction force = .33 ( 3.6 * 9.81 * cos 35 + F) = 9.55+.33 F
tat has to balance component of weight down slope
9.55 + .33 F = 3.6 * 9.81 * sin 35 = 20.3
.33 F = 10.7
F = 32.4 N
friction force = .33 ( 3.6 * 9.81 * cos 35 + F) = 9.55+.33 F
tat has to balance component of weight down slope
9.55 + .33 F = 3.6 * 9.81 * sin 35 = 20.3
.33 F = 10.7
F = 32.4 N
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.