Question

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Here's One more question ~

A crate of mass m is pulled with a force F along a fixed right angled horizontal through as in the figure. The Coefficient of kinetic friction between the crate and the trough is μ . Find the value of force F required to pull it along the trough with constant velocity.

Picture is attached in question.

Thanks for Answering ~

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Answer 1

The crate is moving at constant velocity when the forces acting on it are

balanced.

• The value of the force F required to pull the crate with constant velocity is; $\underline{F = \sqrt{2} \cdot \mu \cdot m\cdot g}$

Reasons:

Mass of the crate = m

Cross section of through = Right angled

Orientation (inclination) of the through to horizontal = 45°

Coefficient of kinetic friction = μ

Required:

The value of the required to pull the crate along the through at constant

velocity.

Solution:

When the through is moving at constant velocity, we have;

Friction force acting on crate = Force pulling the crate

Friction force = Normal reaction × Coefficient of kinetic friction

Normal reaction on an inclined plane = $\mathbf{F_N}$

Each side of the through gives a normal reaction.

The vertical component of the normal reaction on each side of the through

is therefore;

• $F_N$·j = $\mathbf{F_N}$ × sin(θ)

The sum of the vertical component = $F_N$·j + $F_N$·j = 2·$F_N$·j = 2·$F_N$×sin(θ)

The sum of the vertical component of the normal reactions = The weight of the crate

Therefore;

2·$F_N$×sin(θ) = m·g

θ = 45°

Therefore;

2·$F_N$×sin(45°) = m·g

$\displaystyle sin(45^{\circ}) = \mathbf{\frac{\sqrt{2} }{2}}$

Therefore;

$\displaystyle 2 \cdot F_N \cdot sin(45^{\circ}) = 2 \cdot F_N \times \frac{\sqrt{2} }{2} = \sqrt{2} \cdot F_N$

$\displaystyle F_N = \mathbf{ \frac{m \cdot g}{\sqrt{2} }}$

Which gives;

$\displaystyle Force \ required, \ F = Sum \of \ friction \ forces \ = 2 \times F_N \times \mu = \mathbf{ 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu}$

$\displaystyle Force \ required, \ F = 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu = \mathbf{ \sqrt{2} \cdot \mu \cdot m \cdot g}$

• Force required to pull the crate at constant velocity, F = √2·μ·m·g

Learn more about force of friction here:

brainly.com/question/6561298

Answer 2

Answer:

$f = \sqrt{2}μmg \\ 2N \sin(45°) = mg \\ N = \frac{mg}{ \sqrt{2} } \\ f = 2μN \\ 2μ \frac{mg}{ \sqrt{2} } = \sqrt{2} mg.$

hope helpful <3"