Calculating Forces for Angles of 30, 45 & 60

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The discussion focuses on calculating the weight, normal force, and net force for a block at angles of 30, 45, and 60 degrees. The weight of the block is confirmed to be 20 kg, but it must be converted to Newtons using the formula weight = mass * gravity. Participants clarify the correct application of trigonometric functions for calculating normal and net forces, emphasizing the importance of drawing a diagram to visualize the forces. The correct formulas are established, with normal force calculated as weight * cosine of the angle and net force as weight * sine of the angle. The conversation concludes with a confirmation that the calculations are now correct and an acknowledgment of the initial confusion.
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Homework Statement


angles of 30, 45, 60. calculate weight of block, the normal force, and net force for each angle separately.

Homework Equations


normal force m*g*sin theta
net force m*g*cos theta



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...
 
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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...?
 
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.
 
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

?
 
  • #10
is that correct?
 
  • #11
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!
 
  • #12
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!
 
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