Difficult Force problem

  • #1
Hi,

This is just a force and motion problem with three unknowns, pulled out of Haliday Resnick Walker (7th Edition), Chpt. 6.

17) An initially stationary box of sand is to be pulled across a floor by means of a cable in which the the tension should not exceed 1100 N. The coefficiient of static friction between the box and the floor is 0.35.

(a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand

(b) What is the weight of the sand and box in that situation?

Seems easy enough...but I can't seem to set up all the necessary equations to solve this problem.

Here's what I have so far.

Ok pretty staightforward, determine if the system has acceleration, I reasoned that it doesn't, therefore,

(sum of the forces as a vector) = 0

-----------------------------------------------------------------------
Ok, Now I sum the forces in x.

(sum of the forces in x) = 0

(sum of the forces in x) = T*cos(theta) - f_s

(0)= F*cos(theta) - f_s

f_s= F*cos(theta)

(u*N) = F*cos(theta)

----------------------------------------
And then sum them in the y.

(sum of the forces in y) = 0

(sum of the forces in y) = T*sin(theta) + N - W

(0) = T*sin(theta) + N - W

N = -T*sin(theta) + W

----------------------------------------

W = m*g

therefore,

Ok, here are my equations.

u*N = F*cos(theta)

N = -T*sin(theta) + m*g

Now from here, I see I have two equations -- three unknowns: theta, m, and N.

I don't know what's my third equation . . .

or

How, of these two equations, can I get one of the unknowns to cancel itself out and solve for at least one of my variables.

I thought about using "tan (theta)" however

tan(theta) = (T*sin(theta))/(T*cos(theta))

Which reduces back to itself...

And when I did this substituting (T*sin(theta) and (T*cos(theta)) from my earlier two equations, I still can't seem to be able to solve for at least one of the unknowns....

Any help would be most appreciated, Thank you! :)


***NOTE:

u = coefficient of static friction (0.35)
N = Normal Force
W = weight
T = Tension (1100 N)
f_s = force of static friction
 

Answers and Replies

  • #2
0rthodontist
Science Advisor
1,230
0
You are correct, those two equations are all you need. But look at what they mean: they are the equations that hold when the block is moving at a constant speed, when pulled at a given angle and when having a given mass. There are many angles and masses that achieve this, and what you want is the one where m is maximum. If you simplify you'll have (expression involving m and some constants) = (cos theta + mu sin theta). In order to maximize m you'll need to maximize (cos theta + mu sin theta), which can be done using the sine angle addition formula and some algebra.
 
Last edited:
  • #3
SGT
0rthodontist said:
You are correct, those two equations are all you need. But look at what they mean: they are the equations that hold when the block is moving at a constant speed, when pulled at a given angle and when having a given mass. There are many angles and masses that achieve this, and what you want is the one where m is maximum. If you simplify you'll have (expression involving m and some constants) = (cos theta + mu sin theta). In order to maximize m you'll need to maximize (cos theta + mu sin theta), which can be done using the sine angle addition formula and some algebra.
Or you could take the derivative of m relative to theta and make it equal to zero, to find the point of maximum.
 
  • #4
0rthodontist
Science Advisor
1,230
0
You could, but that would give you (mu cos theta - sin theta) = 0 which you'd have to solve using the cosine angle addition formula and some algebra.
 
  • #5
SGT
0rthodontist said:
You could, but that would give you (mu cos theta - sin theta) = 0 which you'd have to solve using the cosine angle addition formula and some algebra.
The solution of that equation is simply
tan theta = mu
No need of any complicated algebra.
 
  • #6
9
0
Solved

The solution of that equation is simply
tan theta = mu
No need of any complicated algebra.

I agree, taking the derivative is easier. dm / d(theta), when equated to zero, gives tan(theta) = [tex]\mu[/tex], so [tex]\mu[/tex] = atan .35, giving [tex]\theta[/tex] = 19.3 degrees.

Substituting in the expression for mass, the value of the highest mass that can be pulled comes out to be 340 kg.
 
  • #7
I initially solved theta for this by taking the derivative of accelleration in the horizontal direction. Although the problem states that initially the box is stationary I assumed that it began moving since it was "to be pulled" across the floor. Am I wrong in taking this alternative approach?
 

Related Threads on Difficult Force problem

Replies
2
Views
2K
Replies
1
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
14
Views
3K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
7
Views
1K
Top