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Hoochie [10]
3 years ago
5

Why is it more difficult to lean over and push a heavy box across the floor than it is to attach a rope and pull the box at the

same angle?

Physics
1 answer:
melisa1 [442]3 years ago
8 0

Answer:

Case 1: <u>Pushing</u> Diagram 1

Leaning over and Pushing the heavy box from the floor, the push will be divided in to two parts, one is horizontal that can help the box move, and one is vertically downwards, which increases the downward force of the heavy object (an addition to the gravity) and thus increases friction, making it very hard to push.  When you push at certain angle, you are exhibiting two forces as shown in diagram 1.

  1. Horizontal force acting along the plane.
  2. Vertical force downward perpendicular to the surface.

Case 2: <u>Pulling</u> Diagram 2

Pulling on a rope similar object at the same angle, the pull can be divided into two parts, one is horizontal that can help the box move, and one is vertically upwards, which decreases the downwards force of the box (a subtraction in the gravity) and thus decreases friction, making it very easy to pull. When you pull at a certain angle, you are exhibiting two forces as shown in diagram 2.

  1. Horizontal force acting along the plane.
  2. Vertical force upward perpendicular to the surface.

So, in the case of pushing, it adds an extra weight on the object, which results in difficulty to push that object at the same angle.  In case of pulling, the upward perpendicular force, it tries to lift the  object upward and divided the weight partially. Thus making it easier to move the object at same angle.

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What is the resistance of 1.0 m of a solid cylindrical metal cable having a diameter of 0.40 inches and a resistivity of 1.68 ×
Dahasolnce [82]

Given;

redistivity =1.68/100000000W.m

diameter =0.40in

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8 0
3 years ago
A block is at rest. The coefficients of static and kinetic friction are s = 0.81 and k = 0.69, respectively. The acceleration of
lys-0071 [83]

Answer:

Explanation:

a ) The angle required = angle of repose = θ

Tanθ = .81

θ = 39⁰

b ) when angle of incline θ = 44

Net force on the block = mg sinθ - μ mg cosθ where μ is coefficient of kinetic friction

acceleration = gsinθ - μ g cosθ

= 9.8 ( sin44 - μ cos44 )

= 9.8 ( .695 - .69 x .72 )

= 9.8 ( .695 - .497 )

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7 0
3 years ago
A 4.1 object is lifted to a height of 4.5m above the surface of Earth.What is the potential energy of the object due to gravity?
liberstina [14]

Answer:

P = 180.81 J

Explanation:

Given that,

Mass of a object, m = 4.1 kg

It is lifted to a height of 4.5 m

We need to find the potential energy of the object due to gravity. It is given by the formula as follows :

P = mgh Where g is acceleration due to gravity

P = 4.1 kg × 9.8 m/s² × 4.5 m

P = 180.81 J

Hence, the potential energy is 180.81 J.

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3 years ago
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