In this question use g = 9.8 m.s^-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. 13 (a) The boy applies a horizontal force of 150 N. Show that the crate remains stationary. 13 (b) 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 Further Forces & Newton's Laws: In this question use g = 9.8 m.s

In this question use g = 9.8 m.s^-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

In this question use g = 9.8 m.s^-2 A boy attempts to move a wooden crate of mass 20 kg along horizontal ground. The coefficient of friction between the crate and t... show full transcript

Worked Solution & Example Answer:In this question use g = 9.8 m.s^-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

Show that the crate remains stationary.

96%

114 rated

Answer

To determine if the crate remains stationary, we need to compare the applied force to the maximum static friction force.

Calculate the weight of the crate. The weight (W) can be calculated using the formula:

$W = mg = 20 ext{ kg} imes 9.8 ext{ m.s}^{-2} = 196 ext{ N}$
Calculate the maximum static friction. The maximum static friction (F_{max}) is given by:

$F_{max} = ext{friction coefficient} imes W = 0.85 imes 196 ext{ N} = 166.6 ext{ N}$
Compare the applied force with the maximum static friction. The boy applies a horizontal force of 150 N, and since:

$150 ext{ N} < 166.6 ext{ N}$

this indicates that the crate will not move and will remain stationary.

Step 2

Determine whether the crate remains stationary.

99%

104 rated

Answer

To analyze the situation with the force applied at an angle, we can break down the 150 N force into its horizontal and vertical components.

Resolve the applied force into components.
- Horizontal component:
$F_{horizontal} = 150 imes ext{cos}(15^ ext{o}) \\ = 150 imes 0.9659 \\ = 144.9 ext{ N}$
- Vertical component:
$F_{vertical} = 150 imes ext{sin}(15^ ext{o}) \\ = 150 imes 0.2588 \\ = 38.82 ext{ N}$
Calculate the normal force (R). The normal force is affected by the vertical component of the applied force:

$R = W - F_{vertical} = 196 ext{ N} - 38.82 ext{ N} = 157.18 ext{ N}$
Determine the maximum static friction with the new normal force.

$F_{max} = ext{friction coefficient} imes R = 0.85 imes 157.18 ext{ N} = 133.6 ext{ N}$
Compare the horizontal component of the applied force with the maximum static friction. Since:

$144.9 ext{ N} > 133.6 ext{ N}$

this means that the force exceeds the maximum static friction, so the crate will begin to move.

In this question use g = 9.8 m.s^-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 m.s^-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

Show that the crate remains stationary.

Determine whether the crate remains stationary.

Join the A-Level students using SimpleStudy...