Inclined Plane Normal Force

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golgiapparatus88

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This problem is really bugging me. I was helping my brother do his high school physics online homework and I can not figure this one out..I really thought I understood this stuff to..

A block with a mass of 20 kg is held in equilibrium on an incline of angle θ = 30.0° by the horizontal force, F, as shown in Figure 4-31. Find the magnitude of F.

Force F is just running parallel to the inclined plane pushing on the block.


Here's how I solved it (wrong)

The normal force is the Y-component of gravitational force. Fn would be mgcos30 which is 173N

The F1 pushing the block is only in the x-direction so that has to equal the x-component of gravity since a=0. mgsin30 is 100N

The answer key says 113N and 227N for the answers respectively. Can someone explain?

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I'm no Physics genius, so this is a very confusing problem. But I'll try my best to use what I know to help explain this.

I did a google search and found the figure to your problem:
10mk4gg.png


First off, it's important to notice that the force being applied isn't parallel to the incline (like you assumed). Instead, only a component of this unknown force acts in the parallel direction. Because the problem states that the object is in equilibrium, naturally I assumed that the component of this force was equivalent to the downward gravitational force acting on the incline.

Solving for unknown component:

Fnet = mgsin(θ) - x
x = mgsin(θ)
x = (20)(9.8)(30)
x = 98 N (I'll call this Fx or Adjacent Side)

This is just one of the sides of the triangle. We need to solve for the opposite side.

Using the trig. function: tan(θ) = Opposite/Adjacent, I was able to solve for the opposite side of the triangle:

tan(θ) x Adjacent = Opposite
tan(30) x 98 N = 56.6 N (I'll call this Fy or Opposite Side)

Then, we can use the Pythageorean Theorem to find the Hypothenuse. The hypothenuse equals the force we're looking for.
What we solved for earlier were just the components of this force.

So: A² + B² = C²

sqrt[(56.6 N)² + (98)²] = C
113.17 N = C

For part B, two forces contribute to the normal force. Namely the upward force of the incline and the Fy component of the applied force.
No need to solve for Fy again since we did that already (above).

The normal force of the incline is: mgcos(θ) or 169.7 N

Now simply adding the two forces, we get a net normal force of: 169.7 N + 56.6 N = 226.3 N
 

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