Motion and Force

  • Thread starter Priscilla
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  • #1
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Could someone please help me take a look at my work?
Especially Question 5, I am not sure what 5a is asking, and I don't know if I could calculate "Us" using the way I did.

Homework Statement





Homework Equations





The Attempt at a Solution

 

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  • #2
Lok
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Damn it, I forgot...
What is the international measuring unit for weight and Force? Put the values in their right place... (5 Q answer) ... It is not a necessity but think of the children who read it like that and turn evil.

Answer for 5 seems ok but I cannot find a 5a question?
 
  • #3
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Sorry, I mean 5c.
 
  • #4
jhae2.718
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For 5c, what force does the conveyor belt exert on the package? Look at your free body diagram...
 
  • #5
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I calculate the "x" and "y", do I just add them together?
 
  • #6
jhae2.718
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Edit: Okay, so you're basically calling the frictional force y and the normal force x.

These forces act in reaction to the motion of the package, so the total reaction force [tex]\vec{R}[/tex] would be the vector sum of x and y, or [tex]\vec{R}=x\hat{i}+y\hat{j}[/tex]. The magnitude of a vector, [tex]|\vec{R}|=R=\sqrt{R_x^2+R_y^2}[/tex]. (Are you familiar with unit vector notation?)

However, based on the problem statement, I think they just want the reaction force in the [tex]\hat{j}[/tex] direction, which would be...?
 
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  • #7
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Does that mean Fn is the answer for this question?
 
  • #8
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Also, did I do the other questions correctly?
 
  • #9
jhae2.718
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That's what I think they're after; based on the equation it seems like it is trying to reinforce the relationship between the normal force and friction.

(I.e. first you find the normal force, which you then use, along with ∑F=ma to find μ.)

What kind of physics course is this?

Edit: I only briefly looked through your work, but you seemed to do everything correctly.
 
  • #10
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The way that I find "Us" is not correct?
There is no velocity and time, how can I find acceleration?

The course name is Physical principles of food strucuture and functionality. I have no idea why we are doing physics!
 
  • #11
jhae2.718
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The way that I find "Us" is not correct?
There is no velocity and time, how can I find acceleration?
No, you found μs correctly.

Since the package is at rest, the net acceleration is equal to zero, so ∑F=ma=0 along the incline.

If we have ∑Fx=mgsinθ-Fs=0 and ∑Fy=N-mgcosθ=0, we can solve them separetly (as you did) to get:
mgsinθ=Fs
N=mgcosθ

If we assume that static friction is at its maximum, then FssN.

Using these three equations, we find that μs=(mgsinθ)/(mgcosθ)=tanθ
 
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  • #12
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What about other questions? Did I do them correctly?
 
  • #13
jhae2.718
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Briefly reading through it, you seemed to have done everything correctly.

Edit: Actually, on 2, you just found the vector sum of the two forces given. You need to find a third vector whose magnitude should be equal to the magnitude of the first two, but opposite to the direction of the sum of the first two.

You have three forces, F1, F2, and F3. You know F1 and F2, and you know that ∑F=0, so you need to break the forces into their component form and find what F3 needs to be to make ∑Fx=0 and ∑Fy=0.
 
  • #14
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Thank you so much!

By the way, the course is under Nutrition and Food Science program....
 
  • #15
jhae2.718
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See my edited post; I missed a mistake you made on 2.
 
  • #17
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So the answer will be -25N with an angle to the horizontal of 1.5 degree?
 
  • #18
jhae2.718
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You have the resultant force of the first two forces already , by the way. The third force is this force with the components negated.

I just want to make sure you understand the process of summing the forces to find the unknown force.

The magnitude is always positive. (The magnitude of a vector [tex]\vec{R}=\langle r_x,r_y \rangle[/tex] is defined as [tex]R=\sqrt{r_x^2+r_y^2}[/tex]. This quantity is always positive for real numbers. The individual components of the third force's vector are both negative. The magnitude is positive.

Your angle should come out to what you had before, but this force is in the fourth quadrant, so you need to add 180 deg. to the result of the arctangent.
 
  • #19
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o~ I got it! Thanks alot!
 
  • #20
jhae2.718
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So the third force should be 25N at 181.5 degrees measured from the x-axis.

No problem.
 

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