Solving Incline Ramp Problem: Find Normal Force

[1irishman]

Homework Statement

Imagine a ramp on a 30deg incline with a 1kg mass on top of it. mu kinetic coefficient is .25 between the mass and the ramp.
A person applies a horizontal force of 10N to the 1kg mass on the incline ramp opposite to the direction of acceleration.
What is the normal force acting on the mass?

Homework Equations

Ff=uFn
Trig functions sin and cosine i think but not sure.

The Attempt at a Solution

I figured the normal force to be 9.8cos 30deg = 8.5N
I figured the force along the ramp in the direction of the acceleration to be 9.8sin 30deg=4.9N
I got the force friction to be .25*8.5=2.1N
I'm lost ...don't know what i am doing wrong because they gave an answer of 13.5N for the normal force acting on the mass. Please help?

 

[hage567]

1. Homework Statement
Imagine a ramp on a 30deg incline with a 1kg mass on top of it. mu kinetic coefficient is .25 between the mass and the ramp.
A person applies a horizontal force of 10N to the 1kg mass on the incline ramp opposite to the direction of acceleration.
What is the normal force acting on the mass?

2. Homework Equations
Ff=uFn
Trig functions sin and cosine i think but not sure.

I figured the normal force to be 9.8cos 30deg = 8.5N

But this is only true if gravity is the only other force. You have the applied force of 10N, which will change the normal force.

I figured the force along the ramp in the direction of the acceleration to be 9.8sin 30deg=4.9N
I got the force friction to be .25*8.5=2.1N

The normal force will depend on the other forces present in the same direction (so in this case perpendicular to the surface of the incline). You must sum up the forces acting in this direction only. Does the frictional force have any component in that direction?

 

[1irishman]

Okay, but the 10N applied force...does that 10N get added to 8.5N? I don't know how to calculate the components with this horizontal force of 10N.

 

[hage567]

Okay, but the 10N applied force...does that 10N get added to 8.5N?

Some of it will.
The horizontal force is at 30 degree angle with respect to the surface of the incline, so some of it contributes to the x direction, some to the y. Calculate it the same way you find any other components of a force. Draw out a diagram. Use trig to find the component that acts in the y direction (still taking this to be perpendicular to the incline).

 

[1irishman]

10sin30=5 so 8.5 + 5 = 13.5 for normal force...right?

 

[1irishman]

5N of the 10N horizontal force contributes to the y direction right?

 

[hage567]

Yes, that's right.