A wooden crate of mass 20kg is pulled in a straight line along a rough horizontal floor using a handle attached to the crate. The handle is inclined at an angle $\alpha$ to the floor, as shown in Figure 1, where $\tan \alpha = \frac{3}{4}$. The tension in the handle is 40N. The coefficient of friction between the crate and the floor is 0.14 The crate is modelled as a particle and the handle is modelled as a light rod. Using the model, a) find the acceleration of the crate. The crate is now pushed along the same floor using the handle. The handle is again inclined at the same angle to the floor, and the thrust in the handle is 40N as shown in Figure 2 below. b) Explain briefly why the acceleration of the crate would now be less than the acceleration of the crate found in part (a).

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A wooden crate of mass 20kg is pulled in a straight line along a rough horizontal floor using a handle attached to the crate - Edexcel - A-Level Maths Mechanics - Question 7 - 2018 - Paper 1

Question 7

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A wooden crate of mass 20kg is pulled in a straight line along a rough horizontal floor using a handle attached to the crate. The handle is inclined at an angle $\a... show full transcript

Worked Solution & Example Answer:A wooden crate of mass 20kg is pulled in a straight line along a rough horizontal floor using a handle attached to the crate - Edexcel - A-Level Maths Mechanics - Question 7 - 2018 - Paper 1

Step 1

Resolve vertically

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Answer

To find the acceleration of the crate, we start by resolving the forces acting on it vertically. The equation can be set up as:

$R + 40 \sin \alpha = 20g$

Here, R is the normal reaction force, $\alpha$ is the angle, and g is the acceleration due to gravity (approximately 9.81 m/s²). With $\tan \alpha = \frac{3}{4}$ , we find:

$\sin \alpha = \frac{3}{5} \quad \text{and} \quad \cos \alpha = \frac{4}{5}$

Substituting these values into the equation gives: $R + 40 \times \frac{3}{5} = 20 \times 9.81$

Step 2

Resolve horizontally

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Answer

Next, we resolve the forces horizontally. The tension and the friction act on the crate, and we can express this as:

$40 \cos \alpha - F = 20a$

Here, F is the frictional force and can be calculated using the coefficient of friction (0.14):

$F = 0.14R$

Step 3

Substituting to find acceleration

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Combining the previous equations, we can substitute F into our horizontal equation. From the vertical resolution, we can deduce:

$R = 20g - 40 \sin \alpha$

Substituting this R into the equation for F gives us:

$F = 0.14(20g - 40 \times \frac{3}{5})$

Finally, substituting F back into the horizontal equation allows us to solve for the acceleration 'a'. The final calculation yields:

$a = 0.396 \text{ or } 0.40 \, (m/s^2)$

Step 4

Explain the effect of pushing

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When the crate is pushed instead of pulled, the normal reaction force R increases, which subsequently increases the available frictional force F. However, increasing F leads to a decrease in acceleration a. This can be explained by considering that a larger frictional force opposes the motion. Thus, despite the thrust applied, the net force acting on the crate is reduced, resulting in a smaller acceleration compared to when it was pulled.

A wooden crate of mass 20kg is pulled in a straight line along a rough horizontal floor using a handle attached to the crate - Edexcel - A-Level Maths Mechanics - Question 7 - 2018 - Paper 1

Worked Solution & Example Answer:A wooden crate of mass 20kg is pulled in a straight line along a rough horizontal floor using a handle attached to the crate - Edexcel - A-Level Maths Mechanics - Question 7 - 2018 - Paper 1

Resolve vertically

Resolve horizontally

Substituting to find acceleration

Explain the effect of pushing

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