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

[get_physical]

Homework Statement

10kg block on a inclined plane (32degrees) (frictionless) and an applied force 120N horizontally. What is acceleration?

Homework Equations

Fnet = ma

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

 

[grzz]

... From applied force, in the same plane, we have xcos32=120 therefore, x = 141.5...

Explain how you got xcos32=120

 

[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

 

[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.

 

[get_physical]

Let me think...

 

[grzz]

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

 

[get_physical]

Oh I get it now! Thank you!