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y A217-kg crate rests on asul g 0 ASSGT6pd o "ol iON CAPA pushing aarx Chegg Study I Guided Solut AGS Kg Crate's 2 Messages Courses Help Logout Connor ) Willson (Student -section: 11515) 2181 PHYS 0174 Basic Physics for Science and Engineering, CRN 11436 O Taner Notes è Evaluate-Feedback Print Course Contents... HW3 Pushing crate down , not the incline) Is required to start the A 237-kg crate rests on a surface that is inclined above the horizontal at an angle of 18.8, A horizontal force (magnitude S05 N and paraliel to the ground down the incline. What is the coefficient of static friction between the crate and the incline? crate moving 2297 Submit Answer Tries 3/20 Previous Tries send Fed Post DiscussionExplanation / Answer
The forces acting parallel to the incline are the component of the crate’s weight (mgsin), the applied force (Fcos), and the friction force (- n). Therefore:
F(x) = 0 = mgsin + Fcos - n
= [mgsin + Fcos] / n--------------------->(1)
the forces acting perpendicular (in “y”) to the incline; are the normal force (n), the component of the crate’s weight perpendicular to the incline (mgcos), and the vertical component of the applied force (Fsin) , which tends to lift the crate up reducing the normal force making it upward:
F(y) = 0 = n - mgcos + Fsin
n = mgcos - Fsin-------------------------->(2)
Substituting (2) into (1), we get:
= [mgsin + Fcos] / [mgcos - Fsin]
= [(237kg)(9.80m/s²)sin18.8° + 505Ncos18.8°] / [(237kg)(9.80m/s²)cos18.8° - 505Nsin18.8°]
= 0.602
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