## calculating normal force and net force of 20 kg mass going down inclined plane at...

1. The problem statement, all variables and given/known data
angles of 30, 45, 60. calculate weight of block, the normal force, and net force for each angle separately.

2. Relevant equations
normal force m*g*sin theta
net force m*g*cos theta

3. The attempt at a solution
weight of block is just 20 kg, right?

and so for 30 degrees: normal force= 20 kg* 9.81 m/s/s * sin 30, right? and net force=20 kg * 9.81 m/s/s * cos 30...
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study
 No. The weight of an object is measured in lbs or N. kg is a unit of mass. Weight in Newtons = mass in kg*acceleration due to gravity in m/s^2. Draw a diagram showing all your forces.
 wait....my next question then asks how does each force value change as the angle changes and why? in this case, wouldn't the weight be constant...?

## calculating normal force and net force of 20 kg mass going down inclined plane at...

You're absolutely right. (weight is constant). Your calculations are not, however, correct.
 ok, then can you please tell me what i did wrong and how to calculate them correctly? thanks.
 Start with a diagram and label your forces, you will see physically how to apply the trig functions. http://www.cheops-pyramide.ch/khufu-...ined-plane.gif here is a good diagram
 ugh. i'm still confused.
 Gm is the weight, N is the normal force. the Alpha sign is the angle. The same angle of the incline. Do you see that a triangle has formed? You can apply the trig functions of sine and cosine.
 so normal force= 196.2 N * cosine 30=170N, which is what I had, and net force= 196.2 N *sin 30=98.1 ?
 is that correct?
 You had normal force = 196.2sin30 which is not correct. Now your calculations look correct. If there was friction then the net force would be the force going down the incline subtracted from the friction force(going up the incline). But everything looks good! great job!
 right, i did have it like that and just now after i looked at my notes, i realized i accidently switched the two formulas. thanks for your help. i really appreciate it!