6. 0.6/9 points | Previous Answers My Notes A block weighing 72.5 N rests on a p

Question

6. 0.6/9 points | Previous Answers My Notes A block weighing 72.5 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, 0.385 and 0.156 (a) What is the minimum value of F that will prevent the block from slipping down the plane? 181N Enter a number (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) F cos + (mg cos + F sin ) = mg sin
F (cos + sin ) = mg (sin - cos )
F (cos 25 + 0.35 sin 25) = 72.5 (sin 35- 0.385*cos 35)
=-7.638

(b) to start to move up the plane the force F must overcome static friction and the force of gravity downward
F=72.5*(sin25+0.385*cos25)
=55.937

(c)once moving then kinetic friction will be present and F can be reduced
F=72.5*(sin25+0.156*cos25)  
= 40.8901

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