In this question use $g = 9.8 \, ext{ms}^{-2}$ A boy attempts to move a wooden crate of mass 20 kg along horizontal ground. The coefficient of friction between the crate and the ground is 0.85. The boy applies a horizontal force of 150 N. Show that the crate remains stationary. Instead, the boy uses a handle to pull the crate forward. He exerts a force of 150 N, at an angle of 15° above the horizontal, as shown in the diagram. Determine whether the crate remains stationary. Fully justify your answer.

A-Level AQA Maths Mechanics Forces: In this question use $g = 9.8 \,

In this question use $g = 9.8 \, ext{ms}^{-2}$ A boy attempts to move a wooden crate of mass 20 kg along horizontal ground - AQA - A-Level Maths Mechanics - Question 13 - 2018 - Paper 2

Question 13

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In this question use $g = 9.8 \, ext{ms}^{-2}$ A boy attempts to move a wooden crate of mass 20 kg along horizontal ground. The coefficient of friction between the... show full transcript

Worked Solution & Example Answer:In this question use $g = 9.8 \, ext{ms}^{-2}$ A boy attempts to move a wooden crate of mass 20 kg along horizontal ground - AQA - A-Level Maths Mechanics - Question 13 - 2018 - Paper 2

Step 1

The boy applies a horizontal force of 150 N. Show that the crate remains stationary.

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Answer

To determine whether the crate remains stationary, we can calculate the maximum friction force using the formula:

$F_{max} = \, ext{μmg}$

Where:

μ (coefficient of friction) = 0.85
m (mass of the crate) = 20 kg
g (acceleration due to gravity) = 9.8 m/s²

Calculating the maximum friction:

$F_{max} = 0.85 \times 20 \times 9.8 = 166.6 \, ext{N}$

Since the horizontal force applied by the boy is 150 N, we compare the two:

Applied force = 150 N
Maximum friction force = 166.6 N

Since 150 N is less than 166.6 N, the crate does not move and remains stationary.

Step 2

Determine whether the crate remains stationary.

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Answer

In this part, we resolve the forces acting on the crate. The boy exerts a force of 150 N at an angle of 15° above the horizontal. We can resolve this force into horizontal and vertical components:

Horizontal component:
$F_x = 150 \, ext{N} \cos(15°)$
Vertical component:
$F_y = 150 \, ext{N} \sin(15°)$

Calculating these components:

$F_x = 150 \, ext{N} \cos(15°) \approx 145.0 \, ext{N}$
$F_y = 150 \, ext{N} \sin(15°) \approx 38.8 \, ext{N}$

Next, we can determine the normal force (R) acting on the crate:

The weight of the crate is: $W = mg = 20 \, ext{kg} \times 9.8 \, ext{m/s}^2 = 196 \, ext{N}$

The normal force can be expressed as: $R = W - F_y = 196 \, ext{N} - 38.8 \, ext{N} = 157.2 \, ext{N}$

Now, we can calculate the limiting friction using: $F_{limiting} = \, ext{μR} = 0.85 \times 157.2 \approx 133.6 \, ext{N}$

Finally, we compare the horizontal force with the limiting friction:

Applied horizontal force = 145.0 N
Limiting friction = 133.6 N

Since 145.0 N is greater than 133.6 N, the crate begins to move.

In this question use $g = 9.8 \, ext{ms}^{-2}$ A boy attempts to move a wooden crate of mass 20 kg along horizontal ground - AQA - A-Level Maths Mechanics - Question 13 - 2018 - Paper 2

Worked Solution & Example Answer:In this question use $g = 9.8 \, ext{ms}^{-2}$ A boy attempts to move a wooden crate of mass 20 kg along horizontal ground - AQA - A-Level Maths Mechanics - Question 13 - 2018 - Paper 2

The boy applies a horizontal force of 150 N. Show that the crate remains stationary.

Determine whether the crate remains stationary.

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