General Phys with Calc 1 6. A block weighing 85.0 N rests on a plane inclined at
ID: 1445068 • Letter: G
Question
General Phys with Calc 1
6. A block weighing 85.0 N rests on a plane inclined at 25.0° to the horizontal. A force F is applied to the object at 35.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.354 and 0.156. (a) What is the minimum value of F that will prevent the block from slipping down the plane?
(b) What is the minimum value of F that will start the block moving up the plane?
(c) What value of F will move the block up the plane with constant velocity?
Explanation / Answer
a) perpendicular to the incline
N + Fsin(35-25) - mgcos25 = 0
N = mgcos25 - Fsin10
maximumstatic friction, fmax = us N = 0.354 ( mgcos25 - Fsin10 )
along the plane:
mgsin25 - fmax - Fcos10 = 0
mgsin25 - 0.354 ( mgcos25 - Fsin10 ) - Fcos10 = 0
F ( cos10 - 0.354sin10) = mgsin25 - 0.354mgcos25
F = 85 (sin25 - 0.354cos25) / (cos10 - 0.354sin10)
F = 9.37 N
b) when block will start to move upward then friction will act downward along the incline.
mgsin25 + fmax - Fcos10 = 0
mgsin25 + 0.354 ( mgcos25 - Fsin10 ) - Fcos10 = 0
F ( cos10 + 0.354sin10) = mgsin25 + 0.354mgcos25
F = 85 (sin25 + 0.354cos25) / (cos10 + 0.354sin10)
F = 60.4 N
c) to keep moving, kinetic friction will act.
mgsin25 + fk - Fcos10 = 0
mgsin25 + 0.156 ( mgcos25 - Fsin10 ) - Fcos10 = 0
F ( cos10 + 0.156sin10) = mgsin25 + 0.156mgcos25
F = 85 (sin25 + 0.156cos25) / (cos10 + 0.156sin10)
F = 47.38 N
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