A block weighing 82.5 N rests on a plane inclined at 25.0° to the horizontal.?

Question

A force F is applied to the object at 40.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.342 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?

Answer 1

I will assume that the force F is pushing at an angle that is positive w/r/t the horizontal plane, such that it is pushing the block up the plane and reducing the normal force of the block. If this is incorrect, it is simple to change the sign of the vertical component of the force and recalculate the numbers.

The normal force of the block is a bit tricky since the components of the force F must be decomposed into the part that is perpendicular to the plane and the part that is parallel to the plane.
F subtends an angle w/r/t the plane of 40-25 degrees, or 15 degrees.
The part that is perpendicular and upward is
sin(15)*F
which is part of the normal force on the interface of the plane and the block
adding gravity, the normal force is
cos(25)*m*g-sin(15)*F
733.5-sin(15)*F

and the part that is parallel is
cos(15)*F
the part of gravity resisting the force is
sin(25)*m*g
=342 N
a) the question is asking what value of F will balance the gravity force at the threshold of static friction downward on the plane?

Looking at a FBD
The force upward on the plane plus the static friction oppose the gravitational pull down slope
342=(733.5-sin(15)*F)*.342
+cos(15)*F

(342-733.5.342)/
(cos(15)-sin(15)*.342)=F

F=104 N

b) This is asking what will exactly balance the gravity force at the threshold of static friction upward on the plane
Now the FBD has F upward has to overcome gravity and the static friction force

cos(15)*F=
342+(733.5-sin(15)*F)*.342

F(cos(15)+sin(15)*.342)=
342+733.5*.342

F=(342+733.5*.342)/
(cos(15)+sin(15)*.342)

F=562 N

c) This is asking what will exactly balance the gravity and the kinetic friction

same equation as b, except use u kinetic
F=(342+733.5*.156)/
(cos(15)+sin(15)*.156)

F=453.6 N

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Answer 2

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