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Parents Pricing Home A-Level AQA Maths Mechanics Working with Vectors The three forces F₁, F₂ and F₃ are acting on a particle
The three forces F₁, F₂ and F₃ are acting on a particle - AQA - A-Level Maths Mechanics - Question 13 - 2017 - Paper 2 Question 13
View full question The three forces F₁, F₂ and F₃ are acting on a particle.
F₁ = (25i + 12j) N
F₂ = (-7i - 5j) N
F₃ = (15i - 28j) N
The unit vectors i and j are horizontal and vertic... show full transcript
View marking scheme Worked Solution & Example Answer:The three forces F₁, F₂ and F₃ are acting on a particle - AQA - A-Level Maths Mechanics - Question 13 - 2017 - Paper 2
Find the magnitude of F, giving your answer to three significant figures. Only available for registered users.
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To find the resultant force F, we first sum the individual forces:
F = F 1 + F 2 + F 3 = ( 25 i + 12 j ) + ( − 7 i − 5 j ) + ( 15 i − 28 j ) = ( 25 − 7 + 15 ) i + ( 12 − 5 − 28 ) j = 33 i − 21 j . F = F₁ + F₂ + F₃
= (25i + 12j) + (-7i - 5j) + (15i - 28j)
= (25 - 7 + 15)i + (12 - 5 - 28)j
= 33i - 21j. F = F 1 + F 2 + F 3 = ( 25 i + 12 j ) + ( − 7 i − 5 j ) + ( 15 i − 28 j ) = ( 25 − 7 + 15 ) i + ( 12 − 5 − 28 ) j = 33 i − 21 j . Next, we calculate the magnitude of F using Pythagoras's theorem:
∣ F ∣ = s q r t ( 33 ) 2 + ( − 21 ) 2 = s q r t 1089 + 441 = s q r t 1530 a p p r o x 39.1. |F| = \\sqrt{(33)^2 + (-21)^2} = \\sqrt{1089 + 441} = \\sqrt{1530} \\approx 39.1. ∣ F ∣ = s q r t ( 33 ) 2 + ( − 21 ) 2 = s q r t 1089 + 441 = s q r t 1530 a pp ro x 39.1. Thus, the magnitude of F is approximately 39.1 N, rounded to three significant figures.
Find the acute angle that F makes with the horizontal, giving your answer to the nearest 0.1°. Only available for registered users.
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To find the angle θ that the force F makes with the horizontal axis, we use the tangent function:
a n ( θ ) = opposite adjacent = ∣ − 21 ∣ 33 = 21 33 . an(θ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{|-21|}{33} = \frac{21}{33}. an ( θ ) = adjacent opposite = 33 ∣ − 21∣ = 33 21 . Now, we calculate θ:
θ = tan − 1 ( 21 33 ) ≈ 18.4 ° . θ = \tan^{-1}\left(\frac{21}{33}\right) \approx 18.4°. θ = tan − 1 ( 33 21 ) ≈ 18.4°. Thus, the acute angle that F makes with the horizontal is approximately 18.4°, rounded to the nearest 0.1°.
Find F₄, giving your answer in terms of i and j. Only available for registered users.
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Since the forces are in equilibrium, the sum of all the forces must equal zero:
F 1 + F 2 + F 3 + F 4 = 0. F₁ + F₂ + F₃ + F₄ = 0. F 1 + F 2 + F 3 + F 4 = 0. As we previously calculated:
F 1 + F 2 + F 3 = 33 i − 21 j . F₁ + F₂ + F₃ = 33i - 21j. F 1 + F 2 + F 3 = 33 i − 21 j . Hence, we find F₄ as follows:
F 4 = − ( F 1 + F 2 + F 3 ) = − ( 33 i − 21 j ) = − 33 i + 21 j . F₄ = -(F₁ + F₂ + F₃) = - (33i - 21j) = -33i + 21j. F 4 = − ( F 1 + F 2 + F 3 ) = − ( 33 i − 21 j ) = − 33 i + 21 j . Thus, F₄ = (-33i + 21j) N.
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