Solving a Frictionless Incline Problem: Calculating Normal Force

[ur5pointos2sl]

The problem states:

A horizontal force of 77 N pushes a 12 kg block up a frictionless incline that makes an angle of 60 degrees with the horizontal.

What is the normal force that the incline exerts on the block?

After drawing the FBD I can see there are only 2 forces acting in the Y direction which are Fn and W.

so Fn = mg cos theta = 12 kg ( 9.81 m/s^2) * cos (60 deg) = 58.86 N.

This is a homework problem and when attempting to submit the problem on webassign it keeps telling me it is incorrect. I am not sure where I am going wrong or if there is an error with the problem.

Thanks.

 

[Delphi51]

The 58.86 N looks good for the normal component of the weight.
Now find and add the normal component of the push.

 

[Doc Al]

After drawing the FBD I can see there are only 2 forces acting in the Y direction which are Fn and W.

How are you defining the Y direction? Perpendicular to the surface?

There are three forces acting on the block. What are their directions?

so Fn = mg cos theta = 12 kg ( 9.81 m/s^2) * cos (60 deg) = 58.86 N.

To find Fn you must consider all forces that have a component perpendicular to the surface: All three forces must be considered.

 

[ur5pointos2sl]

How are you defining the Y direction? Perpendicular to the surface?

There are three forces acting on the block. What are their directions?

To find Fn you must consider all forces that have a component perpendicular to the surface: All three forces must be considered.

Yes the positive Y direction is perpendicular to the surface of the incline. The positive X direction is parallel to the force pushing or pulling the block.

The three forces acting on the block are the Normal force Fn(positive y, No x component), The pull or push force F(positive x, No y component), and also the Weight of the block W(has 2 components, one in the x, one in the y).

Therefore, I would think the only force contributing would be the Weights cos 60 component.

The 58.86 N looks good for the normal component of the weight.
Now find and add the normal component of the push.

When you say the normal component of the push. If I am taking the x direction to be parallel to the push then it would have no normal force. Am I misinterpreting the problem?

 

[Doc Al]

When you say the normal component of the push. If I am taking the x direction to be parallel to the push then it would have no normal force. Am I misinterpreting the problem?

The push force is horizontal, not parallel to the surface. So it will have an x and a y component.