# Applied force in horizontal direction on block on a inclined plane.

1. Jan 2, 2014

### get_physical

1. The problem statement, all variables and given/known data
10kg block on a inclined plane (32degrees) (frictionless) and an applied force 120N horizontally. What is acceleration?

2. Relevant equations

Fnet = ma

3. The attempt at a solution

mgsin32 = 51.9N (from mg)
From applied force, in the same plane, we have xcos32=120 therefore, x = 141.5
141.5-51.9=ma
a=9m/s^2 but the answer is 5.0m/s^2

2. Jan 2, 2014

### grzz

Explain how you got xcos32=120

3. Jan 2, 2014

### get_physical

because the horizontal force is 120N, so I need to find the hypotenuse component. the angle in that triangle is also 32, so xcos32 = 120

4. Jan 2, 2014

### grzz

Think carefully.

The horizontal force of 120N has a component along the plane and another component perpendicular to the plane and you need to find the component along the plane.

5. Jan 2, 2014

### get_physical

Let me think...

6. Jan 2, 2014

### grzz

Remember that when a force is resolved into components, the force itself will be what you are calling the 'hypotenuse'.

7. Jan 2, 2014

### get_physical

Oh I get it now! Thank you!