Newton's Laws of motion problem

AI Thread Summary
A block weighing 70.0 N is on a 25.0° incline, with a force F applied at 35.0° to prevent slipping. The coefficients of static and kinetic friction are 0.333 and 0.156, respectively. The user attempts to solve the problem using a Free Body Diagram and Newton's Second Law, but encounters confusion regarding the forces involved. Key discussions highlight the need to consider forces in both the y-direction and x-direction for a complete analysis. Clarification is sought on calculating the maximum static friction and the correct application of the equations.
sepah50
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Homework Statement



A block weighing 70.0 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, 0.333 and 0.156.

What is the minimum value of F that will prevent the block from slipping down the plane?

Homework Equations



\sumFy= Sum of all Forces = Newton's Second Law

The Attempt at a Solution



So in the beginning I draw the Angles and make a Free Body Diagram. I sum up all the forces which is
\sumFy = \eta + Fsin(10) - 70cos(25) = may

may is 0 since block is not moving away from the plane

what I get then is \eta= .173648F - 63.4415

After this point I know I have to find Fsmax which equals \mus\eta. After this part I get lost but I can't find \eta\mus
From what I remember \mu\etas = mg*sin\theta. Hope somebody can help!
 
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Hi sepah50,

sepah50 said:

Homework Statement



A block weighing 70.0 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, 0.333 and 0.156.

What is the minimum value of F that will prevent the block from slipping down the plane?

Homework Equations



\sumFy= Sum of all Forces = Newton's Second Law



The Attempt at a Solution



So in the beginning I draw the Angles and make a Free Body Diagram. I sum up all the forces which is
\sumFy = \eta + Fsin(10) - 70cos(25) = may

may is 0 since block is not moving away from the plane

what I get then is \eta= .173648F - 63.4415

I think you have a couple of sign errors here.

After this point I know I have to find Fsmax which equals \mus\eta. After this part I get lost but I can't find \eta\mus

When you summed up all the forces, you did it only in the y-direction (perpendicular to the plane). What is the sum of the forces in the x-direction (parallel to the plane)?

From what I remember \mu\etas = mg*sin\theta. Hope somebody can help!
 
Thanks! :)
 
thanks
 
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