Solving a Force Problem on a 60° Incline

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A 2Kg mass on a 60° incline is in equilibrium, requiring the calculation of horizontal and normal forces. The user initially calculates gravitational force as 19.62N but is confused about the correct values of 34N for horizontal force and 39N for normal force. The discussion emphasizes the importance of free body diagrams to visualize and resolve forces acting on the mass, which include components along and perpendicular to the incline. It highlights that normal force differs from gravitational force due to the incline's angle. Properly applying trigonometry to these components is essential for accurate calculations.
GWelch
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I know this must be a very simple problem but the answer is puzzling me.

A 2Kg mass is sitting on a 60 degree incline (disregard friction) and i held in equilibrium.
A) What is Horizontl Force (book answer 34N)
B) What is Nomal Force (book answer 39N)

The only force I see acting on the mass is Gravity, so my answer is 2x9.81 equals 19.62N
The Normal Force would also be 19.62N with no Fricition.

What am I missing.

Thanks in advance for your time and trouble.
 
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You seem to have trouble with freebody diagrams and resolving forces. You can't "see" a force unless you draw a properly labelled diagram accounting for each force that acts on the body. The idea is to resolve forces that you 'see' in two mutually perpendicular directions.

Here, the weight has two components--along the plane and perpendicular to it. Note that the block is in equilibrium. The equation of motion therefore reduces to a force balance equation. Does that give you a hint?

Cheers
Vivek
 
Also remember that on an incline, normal force does not equal gravitational force.
 
just draw a free body diagram and include gravity as well as the normal force. If you use simple trig, you can find force normal, which is perpendicular to the surface of the incline.
 
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