Optimizing Friction for Moving Sand: What Angle Should the Cable Be?

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To optimize the pulling of a box of sand using a cable without exceeding a tension of 1100 N, the angle between the cable and the horizontal should be determined. The coefficient of static friction is 0.35, leading to the suggestion that the optimal angle is arctan(µ). The equations provided for normal force and weight need to be rearranged to express weight as a function of angle. The discussion emphasizes the importance of maximizing the mass by considering the tension and angle. Overall, the correct approach involves using the relationship between tension, angle, and friction to find the optimal angle for pulling the sand.
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Question: An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed 1100 N. The coefficient 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?

My solution that gives me the 'incorrect' answer:

Eqn 1--> Tcos(theta) - u(static coefficient)F(normal)=0

Equation 2--> Force Normal + tsin(theta) - mg=0

I can solve for the force normal to be equal to = mg/1.35

For there i assumed that the angle should be 45 degrees (i know..wrong assumption) but have no clue what to do next...

any help would be appreciated
 
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king_naeem said:
Eqn 1--> Tcos(theta) - u(static coefficient)F(normal)=0

Equation 2--> Force Normal + tsin(theta) - mg=0
These two equations are fine. Rearrange them to show the weight (mg) as a function of the angle. Find the angle that maximizes the weight.
 
how would i rearrange the first equation to get it interms of mg?

b/c the i have

Tcos(theta) - u(static coefficient)F(normal)=0

force normal does not equal the force of gravity because the force applied on the string is at an angle so i can't just say 'force of gravity=force normal' can i? :confused: unless I'm wrong yet again...ahh!
 
okay...is this right so far

i get by combining and rearranging the following new equation:

Cos (theta) + Sin(theta)= 0.003121363m

where m is the mass of the object...what should i do next?
 
king_naeem said:
how would i rearrange the first equation to get it interms of mg?

b/c the i have

Tcos(theta) - u(static coefficient)F(normal)=0

force normal does not equal the force of gravity because the force applied on the string is at an angle so i can't just say 'force of gravity=force normal' can i? :confused: unless I'm wrong yet again...ahh!

1.I should like to congratulate DocAl for his 3000 posts.
2.From the first equation express N (the normal force of pressing) in terms of µ,T and theta.Plug it in the second equation and obtain the function m=m(µ,theta,T,g).Since g is constant,for the max of m you must have max tension T (1100 N) and a special angle which is found by the condition of maximum imposed to m seen as a function of theta (considering T as a parameter).My guess is that angle should be arctan (µ).Then insert all numerical data and get the maximized mass.

I believe it's all clear now.
 
dextercioby said:
1.I should like to congratulate DocAl for his 3000 posts.
Thanks, Daniel!
2... My guess is that angle should be arctan (µ).
Good answer (and good "guess"). :wink:
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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