- #1
Tony Zalles
- 22
- 0
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
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