# Friction Problem Homework: Find Angle & Weight of Box

• nothingsus
In summary, to find the angle that maximizes the amount of sand pulled, we can use the maximum static friction equation and the equation for the horizontal component of tension. By setting these two equal and solving for the weight of the sand, we can find the angle that will give us the maximum weight of sand pulled, which will be when the weight is at its maximum.

## Homework Statement

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

## Homework Equations

f_s, max = mu_s*F_n

## The Attempt at a Solution

[/B]

I would like some guidance on my thought process and maybe some hints on how to progress.

Defining up as +y and right as +x
Initially the box isn't moving so sum of F_x and sum of F_y is zero

From the FBD, for the y direction

F_n + T*sin(θ) - F_g = 0

so F_n = F_g - T*sin(θ), so subbing it into the static friction equation

so f_s, max = 0.35*(F_g - T*sin(θ))
= 0.35 (9.8m - 1100sin(θ)) N

From the FBD, for the x direction
T*cos(θ) - f_s = 0
T*cos(θ) = f_s
so f_s = 1100*cos(θ) N

Initially I thought θ = 0° would do, but then I thought maybe to "pull the greatest possible amount of sand", we want to find a θ value that best maximizes the cos part in the horizontal component of tension (so it overcomes the f_s, max) but at the same time best minimizes the sin part in f_s, max.

II have no idea how to do this besides maybe the derivative of f_s, max(θ) , but then I just end up with f's_max = cos(θ) = 0 and that is no help.

Maybe letting f_s = f_s,max could help, but I don't know how to deal with the pesky m or how I would solve for theta.

nothingsus said:

## Homework Statement

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

## Homework Equations

f_s, max = mu_s*F_n

## The Attempt at a Solution

[/B]

I would like some guidance on my thought process and maybe some hints on how to progress.

Defining up as +y and right as +x
Initially the box isn't moving so sum of F_x and sum of F_y is zero

From the FBD, for the y direction

F_n + T*sin(θ) - F_g = 0

so F_n = F_g - T*sin(θ), so subbing it into the static friction equation

so f_s, max = 0.35*(F_g - T*sin(θ))
= 0.35 (9.8m - 1100sin(θ)) N

From the FBD, for the x direction
T*cos(θ) - f_s = 0
T*cos(θ) = f_s
so f_s = 1100*cos(θ) N

Initially I thought θ = 0° would do, but then I thought maybe to "pull the greatest possible amount of sand", we want to find a θ value that best maximizes the cos part in the horizontal component of tension (so it overcomes the f_s, max) but at the same time best minimizes the sin part in f_s, max.

II have no idea how to do this besides maybe the derivative of f_s, max(θ) , but then I just end up with f's_max = cos(θ) = 0 and that is no help.

Maybe letting f_s = f_s,max could help, but I don't know how to deal with the pesky m or how I would solve for theta.
Your FBD looks correct. Write an equation which connects the mass and angle theta. Use maximum friction and maximum normal reaction, since the block is on a horizontal surface.

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nothingsus said:
find a θ value that best maximizes the cos part in the horizontal component of tension
No, you want the value that maximises m. Use the equations you have to find m as a function of theta (f_s eliminated).
cnh1995 said:
maximum normal reaction,
How will maximising the normal reaction help? Wouldn't that be with a negative theta?

haruspex said:
How will maximising the normal reaction help? Wouldn't that be with a negative theta?
Sorry, I meant to say use 'total weight' of the box and sand. In absence of the pulling cable, normal reaction would be equal to the total weight. So I blindly used the term 'normal reaction' here. Normal reaction would be smaller when this cable is pulling the box..

nothingsus said:

## Homework Equations

f_s, max = mu_s*F_n

## The Attempt at a Solution

so f_s, max = 0.35*(F_g - T*sin(θ))

From the FBD, for the x direction
T*cos(θ) = f_s

Maybe letting f_s = f_s,max could help
You derived the formula for the maximum static friction. The friction is horizontal, so it balances with the horizontal component of the tension. Make them equal 0.35*(F_g - T*sin(θ))=T*cos(θ), isolate F_g and find the angle where it is maximum.

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## 1. What is friction and why is it important in this problem?

Friction is the force that opposes motion between two surfaces in contact. In this problem, it is important because it affects the movement of the box and the angle at which it sits on the surface.

## 2. How do I calculate the angle of the box?

The angle of the box can be calculated using the formula: tanθ = μ, where θ is the angle, and μ is the coefficient of friction between the box and the surface.

## 3. How do I determine the weight of the box?

The weight of the box can be determined using the formula: W = mg, where W is the weight, m is the mass of the box, and g is the acceleration due to gravity (9.8 m/s²).

## 4. Can I use any unit for the weight and angle?

It is important to ensure that the units used for weight and angle are consistent throughout the problem. In this case, the weight should be in Newtons (N) and the angle should be in degrees (°).

## 5. How does the coefficient of friction affect the angle and weight of the box?

The coefficient of friction plays a crucial role in determining the angle and weight of the box. A higher coefficient of friction will result in a steeper angle and a heavier weight, while a lower coefficient of friction will result in a lower angle and a lighter weight.