Easy Free Body Diagram Question

In summary, what is wrong with my answer? I've tried just about everything, but its not accepting it. I've tried +N too, even though I think what I have in there should be right.
  • #1
mohabitar
140
0
PEv71.png


Whats wrong with my answer? I've tried just about everything, but its not accepting it. I've tried +N too, even though I think what I have in there should be right.
 
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  • #2
mohabitar said:
Whats wrong with my answer? I've tried just about everything, but its not accepting it. I've tried +N too, even though I think what I have in there should be right.
Look at the definition of the axes below the free body diagram. y is pointed perpendicular (away) from the surface (i.e. normal to the surface). x points parallel to the surface (down the slope).

The question is asking you to find ay. Is the object accelerating into or away from the surface plane?
 
  • #3
Shouldn't up the y-axis be positive and down the y-axis be negative?
 
  • #4
thrill3rnit3 said:
Shouldn't up the y-axis be positive and down the y-axis be negative?
Sure, if you want to define them that way, and if you are given the freedom to do so. But in this problem, the x-axis and y-axis are defined for you. One doesn't have a choice on how to define them, for this particular problem.

[Edit: and yes, in this particular problem, up "the y-axis" does represent a positive y, vy and ay. But my point is that in this particular problem, the y-axis does not point up. It points up and to the right. The problem statement doesn't give you a choice to define it differently.]
 
  • #5
I don't get it. It should just be the sum of all the forces in the y direction, and that's simply the Normal force and the y component of mg, which is mgcos30?
 
  • #6
mohabitar said:
I don't get it. It should just be the sum of all the forces in the y direction, and that's simply the Normal force and the y component of mg, which is mgcos30?
Yes, okay I see what you're saying now.

(My previous line of reasoning is that these two force components add up to something very trivial, and that trivial thing is what is equal to may. But maybe that's not the form of what this online question is asking.)

Try thrill3rnit3's advice. It might be just what this online program thing is looking for. Given how the y-axis as it is defined for you, which term gets the negative sign, the N or the mgcos(30)?
 
  • #7
I think I've tried all possible combinations of signs (+,-) to see if it works, but it wouldn't accept anything. It should be N-mgcos30 then right?
 
  • #8
collinsmark said:
(My previous line of reasoning is that these two force components add up to something very trivial, and that trivial thing is what is equal to may. But maybe that's not the form of what this online question is asking.)

Ah, I see what you're saying. I thought the program thing was asking for an expression for may, not a numerical value.
 
  • #9
Ya it is-it wants an expression not a numerical value
 
  • #10
So which should be positive based from the axes given, N, or mgcos30o?
 
  • #11
mohabitar said:
I think I've tried all possible combinations of signs (+,-) to see if it works, but it wouldn't accept anything. It should be N-mgcos30 then right?
Well, yes I suppose. If I had to use something of that form, "N - mgcos30" would be better.

But if you've tried that and different combinations, maybe my original idea isn't so off the mark (maybe, that is. It depends on the authors of the program). So maybe try this: Besides summing up the individual force components along the y-axis, you already know something about what may is (assuming the object stays on the x-axis and doesn't accelerate away from it). Maybe the program is looking for that, maybe.
 
  • #12
Or maybe the program is barfing because you are inputting "N - mgcos30" instead of "N - mgcos(30)" or something like that.

[Edit Oops. Forgot about the "*" symbols. Make that "N - m*g*cos(30)". My point is that maybe the program is expecting the parentheses around the "30" in the sine function.]
 
  • #13
collinsmark said:
Or maybe the program is barfing because you are inputting "N - mgcos30" instead of "N - mgcos(30)" or something like that.

haha I also think it's an issue with parentheses :rofl:
 
  • #14
If the problem is asking you to find ma_y, where y is the axis perpendicular to the plane, what is the magnitude of a_y? Hint: Think about applying one of Newton's laws in the given y direction.
 

1. What is a free body diagram?

A free body diagram is a visual representation of the forces acting on an object. It is used to analyze the motion of an object and determine the net force acting on it.

2. How do you draw a free body diagram?

To draw a free body diagram, you must first identify the object and all the forces acting on it. Then, draw a dot to represent the object and draw arrows to represent each force, with the length and direction of the arrow indicating the magnitude and direction of the force, respectively.

3. What are the key components of a free body diagram?

The key components of a free body diagram are the object, forces acting on the object, and the direction and magnitude of each force. It is also important to label each force and include a coordinate system to show the direction of motion.

4. Why are free body diagrams important?

Free body diagrams are important because they allow us to clearly visualize and understand the forces acting on an object. They also help us to accurately analyze the motion of an object and determine the net force acting on it.

5. Can you use free body diagrams to solve any type of problem?

Yes, free body diagrams can be used to solve problems involving any type of motion, including linear, circular, and projectile motion. They can also be used to analyze the forces acting on an object at rest or in motion.

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