Asked by Joey
A file cabinet weighing 1500 N is at rest on the floor. The coefficient of static friction between the floor and the cabinet is 0.5. What is the minimum force required to make the file cabinet move horizontally
Answers
Answered by
Henry
Fc = 1500 N @ 0o. = Force of cabinet.
Fp = 150*sin(0) = 0 = Force parallel to floor.
Fv = 1500*cos(0) = 1500 N. = Force perpendicular to floor = Normal.
Fs = u*Fv = 0.5*1500 = 750 N. = Force of
static friction.
Fap-Fp-Fs = m*a.
Fap-0-750 = m*0 = 0
Fap = 750 N. = Force applied.
Fp = 150*sin(0) = 0 = Force parallel to floor.
Fv = 1500*cos(0) = 1500 N. = Force perpendicular to floor = Normal.
Fs = u*Fv = 0.5*1500 = 750 N. = Force of
static friction.
Fap-Fp-Fs = m*a.
Fap-0-750 = m*0 = 0
Fap = 750 N. = Force applied.
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.