Force Required to pull a cylindrical cover

In summary, the total force required to pull the plastic shade cover completely to the right is 5.48 x 10^6 N. An efficient pulley system can be used to reduce the effort by 1/3. In summary, to pull a plastic shade cover of 290 ft in length and 150 ft in breadth, with a thickness of 10 mm and a density of 850 kg/m^3, wrapped around a cylindrical wooden rod of 10 cm diameter and density 650 kg/m^3 at a height of 7 ft from the ground, a total force of 5.48 x 10^6 N is required. To reduce the effort by 1/3, an efficient pulley system
  • #1
vaas44
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1. Homework Statement

A plastic shade cover is almost 290 ft in length and 150 ft in breadth. It is wrapped around a cylindrical wooden rod which is at a height of 7 ft from the ground. The cover is pulled by nylon/plastic ropes to the right to its complete length of 290 ft. Consider the thickness of plastic cover to be 10 mm and density of plastic as 850 kg/m^3. The ropes are pulled by 7 people towards right. Diameter of the wooden rod is 10 cm and density 650 kg/m^3.

How much total force is required to pull the cover completely to the right?

Please give views on an efficient pulley system to reduce the effort by 1/3.


Homework Equations



3. I tried calculating angular acceleration of the cylindrical rod and to further determine the exact force required to pull the cover to its full length.
 

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  • #2
The Attempt at a Solution We can calculate the total force required to pull the cover to its full length by using the formula F = ma, where 'm' is the mass and 'a' is acceleration.Mass of the plastic shade cover = Volume x Density = (290 x 150 x 0.01) m^3 x 850 kg/m^3 = 4.26 x 10^4 kgMass of the wooden rod = Volume x Density = (π x (0.1/2)^2 x 7) m^3 x 650 kg/m^3 = 1.06 x 10^3 kgTotal mass = 4.26 x 10^4 + 1.06 x 10^3 = 4.37 x 10^4 kgAngular Acceleration = 2π(290/150)/7 = 2π(1.966666667) = 12.568075117 rad/s^2Total Force required to pull the cover completely to the right = Mass x Acceleration = 4.37 x 10^4 x 12.568075117 = 5.48 x 10^6 NThus, the total force required to pull the cover completely to the right is 5.48 x 10^6 N. To reduce the effort by 1/3, an efficient pulley system can be used. This pulley system will consist of two or more pulleys arranged in such a way that the force applied to pull the cover is reduced. This is possible because the load is shared by multiple pulleys, thus reducing the effort needed to pull the cover.
 

Related to Force Required to pull a cylindrical cover

1. What is the formula for calculating the force required to pull a cylindrical cover?

The force required to pull a cylindrical cover can be determined using the formula F = μN, where F is the force, μ is the coefficient of friction, and N is the normal force.

2. How does the surface material of the cylindrical cover affect the force required to pull it?

The surface material of the cylindrical cover can affect the force required to pull it by changing the coefficient of friction. Rougher surfaces tend to have higher coefficients of friction, resulting in a greater force required to pull the cover.

3. What factors can impact the coefficient of friction between the cover and its surface?

The coefficient of friction between the cover and its surface can be affected by factors such as the type of material, surface roughness, temperature, and the presence of lubricants or contaminants.

4. Is the force required to pull a cylindrical cover dependent on the size or weight of the cover?

Yes, the force required to pull a cylindrical cover can be influenced by its size and weight. The larger and heavier the cover, the greater the force needed to overcome its inertia and move it.

5. Are there any other factors that should be considered when determining the force required to pull a cylindrical cover?

Other factors that may affect the force required to pull a cylindrical cover include the shape and design of the cover, the angle at which it is being pulled, and the force applied to the cover (e.g. pushing or pulling).

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