Resolving Forces - Revision history

Ellen References at 15:54, 28 November 2019

2019-11-28T15:54:09Z

Ellen References at 17:10, 12 November 2019

2019-11-12T17:10:38Z

NRJC at 20:35, 6 April 2019

2019-04-06T20:35:12Z

NRJC at 17:32, 6 April 2019

2019-04-06T17:32:51Z

NRJC at 17:31, 6 April 2019

2019-04-06T17:31:40Z

NRJC: /* Examples */

2019-02-04T20:47:51Z

‎Examples

NRJC: /* Examples */

2019-02-04T20:47:02Z

‎Examples

NRJC: /* Examples */

2019-02-04T20:36:16Z

‎Examples

NRJC at 20:34, 4 February 2019

2019-02-04T20:34:59Z

NRJC at 18:29, 4 February 2019

2019-02-04T18:29:57Z

← Older revision		Revision as of 15:54, 28 November 2019
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	:[https://www.amazon.co.uk/gp/product/1782945970/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945970&linkCode=as2&tag=nrjc-21&linkId=a120d24dcc7cc7a58192069a3aafc1d2 ''Resolving forces, page 153, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA '']		:[https://www.amazon.co.uk/gp/product/1782945970/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945970&linkCode=as2&tag=nrjc-21&linkId=a120d24dcc7cc7a58192069a3aafc1d2 ''Resolving forces, page 153, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA '']
	:[https://www.amazon.co.uk/gp/product/1471851362/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851362&linkCode=as2&tag=nrjc-21&linkId=7d78d70a2044ee9982dae010c94af92a ''Resolving forces, pages 212-13, GCSE Combined Science Trilogy 2, Hodder, AQA '']		:[https://www.amazon.co.uk/gp/product/1471851362/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851362&linkCode=as2&tag=nrjc-21&linkId=7d78d70a2044ee9982dae010c94af92a ''Resolving forces, pages 212-13, GCSE Combined Science Trilogy 2, Hodder, AQA '']
		+
		+	====Edexcel====
		+
		+	:[https://www.amazon.co.uk/gp/product/1782948163/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782948163&linkCode=as2&tag=nrjc-21&linkId=0fdbfd5dd397d6e24a9dfb250f08587f ''Resolving forces, page 209, GCSE Physics, CGP, Edexcel '']

← Older revision		Revision as of 17:10, 12 November 2019
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	<math>F_x \approx 15N</math>		<math>F_x \approx 15N</math>
	\|}		\|}
		+
		+	===References===
		+	====AQA====
		+
		+	:[https://www.amazon.co.uk/gp/product/178294558X/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=178294558X&linkCode=as2&tag=nrjc-21&linkId=f0dfb66dafcb0c6e9449e7b1a4ae1ac387 ''Resolving (a force), page 54, GCSE Physics; The Revision Guide, CGP, AQA '']
		+	:[https://www.amazon.co.uk/gp/product/1782946403/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782946403&linkCode=as2&tag=nrjc-21&linkId=32a0abb60dff015b15b50e9b1d7b4644 ''Resolving forces, page 133, GCSE Combined Science Trilogy; Physics, CGP, AQA '']
		+	:[https://www.amazon.co.uk/gp/product/1782945970/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1782945970&linkCode=as2&tag=nrjc-21&linkId=a120d24dcc7cc7a58192069a3aafc1d2 ''Resolving forces, page 153, GCSE Physics; The Complete 9-1 Course for AQA, CGP, AQA '']
		+	:[https://www.amazon.co.uk/gp/product/1471851362/ref=as_li_tl?ie=UTF8&camp=1634&creative=6738&creativeASIN=1471851362&linkCode=as2&tag=nrjc-21&linkId=7d78d70a2044ee9982dae010c94af92a ''Resolving forces, pages 212-13, GCSE Combined Science Trilogy 2, Hodder, AQA '']

@@ Line 35: / Line 35: @@
 |-
 | style="height:20px; width:200px; text-align:center;" |The [[diagram]] gives the [[magnitude]] of the [[weight]] as 39N and the [[angle]] of the slope as 22.6°.
-| style="height:20px; width:200px; text-align:center;" |Using the slope as the [[x-axis]] a right angle triangle is drawn with the [[weight]] as the hypotenuse to show the two [[perpendicular]] components. The [[angle]] at the bottom of this [[triangle]] is the same as the [[angle]] of the slope.
+| style="height:20px; width:200px; text-align:center;" |Using the slope as the [[x-axis]] a right angle triangle is drawn with the [[weight]] as the hypotenuse to show the two [[perpendicular]] components. The [[angle]] at the bottom of this triangle is the same as the [[angle]] of the slope.
 |-
 | style="height:20px; width:600px; text-align:center;"; colspan = "2" |

@@ Line 16: / Line 16: @@
 |-
 | style="height:20px; width:200px; text-align:center;" |The [[diagram]] gives the [[magnitude]] of the [[tension]] as 10N and the [[angle]] from the [[horizontal]] as 37°.
-| style="height:20px; width:200px; text-align:center;" |Using the [[horizontal]] as the [[x-axis]] a [[Right Angle Triangle|right angle triangle]] is drawn with the [[tension]] as the hypotenuse to show the two [[perpendicular]] components.
+| style="height:20px; width:200px; text-align:center;" |Using the [[horizontal]] as the [[x-axis]] a right angle triangle is drawn with the [[tension]] as the hypotenuse to show the two [[perpendicular]] components.
 |-
 | style="height:20px; width:600px; text-align:center;"; colspan = "2" |
-The component needed is the [[Adjacent (Trigonometry)|adjacent]] to the [[angle]] in the [[Right Angle Triangle|right angle triangle]] so [[cosine]] should be used.
+The component needed is the adjacent to the [[angle]] in the right angle triangle so cosine should be used.
 <math>F_x = 10\cos(37)</math>
@@ Line 35: / Line 35: @@
 |-
 | style="height:20px; width:200px; text-align:center;" |The [[diagram]] gives the [[magnitude]] of the [[weight]] as 39N and the [[angle]] of the slope as 22.6°.
-| style="height:20px; width:200px; text-align:center;" |Using the slope as the [[x-axis]] a [[Right Angle Triangle|right angle triangle]] is drawn with the [[weight]] as the hypotenuse to show the two [[perpendicular]] components. The [[angle]] at the bottom of this [[triangle]] is the same as the [[angle]] of the slope.
+| style="height:20px; width:200px; text-align:center;" |Using the slope as the [[x-axis]] a right angle triangle is drawn with the [[weight]] as the hypotenuse to show the two [[perpendicular]] components. The [[angle]] at the bottom of this [[triangle]] is the same as the [[angle]] of the slope.
 |-
 | style="height:20px; width:600px; text-align:center;"; colspan = "2" |
-The component needed is the [[Opposite (Trigonometry)|opposite]] to the [[angle]] in the [[Right Angle Triangle|right angle triangle]] so [[sine]] should be used.
+The component needed is the opposite to the [[angle]] in the right angle triangle so sine should be used.
 <math>F_x = 39\sin(22.6)</math>

@@ Line 16: / Line 16: @@
 |-
 | style="height:20px; width:200px; text-align:center;" |The [[diagram]] gives the [[magnitude]] of the [[tension]] as 10N and the [[angle]] from the [[horizontal]] as 37°.
-| style="height:20px; width:200px; text-align:center;" |Using the [[horizontal]] as the [[x-axis]] a [[Right Angle Triangle|right angle triangle]] is drawn with the [[tension]] as the [[hypotenuse]] to show the two [[perpendicular]] components.
+| style="height:20px; width:200px; text-align:center;" |Using the [[horizontal]] as the [[x-axis]] a [[Right Angle Triangle|right angle triangle]] is drawn with the [[tension]] as the hypotenuse to show the two [[perpendicular]] components.
 |-
 | style="height:20px; width:600px; text-align:center;"; colspan = "2" |
@@ Line 35: / Line 35: @@
 |-
 | style="height:20px; width:200px; text-align:center;" |The [[diagram]] gives the [[magnitude]] of the [[weight]] as 39N and the [[angle]] of the slope as 22.6°.
-| style="height:20px; width:200px; text-align:center;" |Using the slope as the [[x-axis]] a [[Right Angle Triangle|right angle triangle]] is drawn with the [[weight]] as the [[hypotenuse]] to show the two [[perpendicular]] components. The [[angle]] at the bottom of this [[triangle]] is the same as the [[angle]] of the slope.
+| style="height:20px; width:200px; text-align:center;" |Using the slope as the [[x-axis]] a [[Right Angle Triangle|right angle triangle]] is drawn with the [[weight]] as the hypotenuse to show the two [[perpendicular]] components. The [[angle]] at the bottom of this [[triangle]] is the same as the [[angle]] of the slope.
 |-
 | style="height:20px; width:600px; text-align:center;"; colspan = "2" |

@@ Line 21: / Line 21: @@
 The component needed is the [[Adjacent (Trigonometry)|adjacent]] to the [[angle]] in the [[Right Angle Triangle|right angle triangle]] so [[cosine]] should be used.
-<math>F_x = 10\cos37</math>
+<math>F_x = 10\cos(37)</math>
 <math>F_x = 7.9863551N</math>
@@ Line 40: / Line 40: @@
 The component needed is the [[Opposite (Trigonometry)|opposite]] to the [[angle]] in the [[Right Angle Triangle|right angle triangle]] so [[sine]] should be used.
-<math>F_x = 39\sin22.6</math>
+<math>F_x = 39\sin(22.6)</math>
 <math>F_x = 14.98751758N</math>

@@ Line 15: / Line 15: @@
 |[[File:ResolvingTension2.png|center|300px]]
 |-
-| style="height:20px; width:200px; text-align:center;" |Text
+| style="height:20px; width:200px; text-align:center;" |The [[diagram]] gives the [[magnitude]] of the [[tension]] as 10N and the [[angle]] from the [[horizontal]] as 37°.
-| style="height:20px; width:200px; text-align:center;" |Text
+| style="height:20px; width:200px; text-align:center;" |Using the [[horizontal]] as the [[x-axis]] a [[Right Angle Triangle|right angle triangle]] is drawn with the [[tension]] as the [[hypotenuse]] to show the two [[perpendicular]] components.
 |}
 {| class="wikitable"
-|+ Calculate the component of the [[tension]] acting against [[friction]].
+|+ Calculate the component of the [[weight]] acting down the slope.
 |-
 |[[File:ResolvingSlope1.png|center|300px]]
 |[[File:ResolvingSlope2.png|center|300px]]
 |-
-| style="height:20px; width:200px; text-align:center;" |Text
+| style="height:20px; width:200px; text-align:center;" |The [[diagram]] gives the [[magnitude]] of the [[weight]] as 39N and the [[angle]] of the slope as 22.6°.
-| style="height:20px; width:200px; text-align:center;" |Text
+| style="height:20px; width:200px; text-align:center;" |Using the slope as the [[x-axis]] a [[Right Angle Triangle|right angle triangle]] is drawn with the [[weight]] as the [[hypotenuse]] to show the two [[perpendicular]] components. The [[angle]] at the bottom of this [[triangle]] is the same as the [[angle]] of the slope.
 |}

← Older revision		Revision as of 20:34, 4 February 2019
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	===Examples===		===Examples===
		+	{\| class="wikitable"
		+	\|+ Calculate the component of the [[tension]] acting against [[friction]].
		+	\|-
		+	\|[[File:ResolveTension1.png\|center\|300px]]
		+	\|[[File:ResolveTension2.png\|center\|300px]]
		+	\|-
		+	\| style="height:20px; width:200px; text-align:center;" \|Text
		+	\| style="height:20px; width:200px; text-align:center;" \|Text
		+	\|}
		+
		+	{\| class="wikitable"
		+	\|+ Calculate the component of the [[tension]] acting against [[friction]].
		+	\|-
		+	\|[[File:ResolveSlope1.png\|center\|300px]]
		+	\|[[File:ResolveSlope2.png\|center\|300px]]
		+	\|-
		+	\| style="height:20px; width:200px; text-align:center;" \|Text
		+	\| style="height:20px; width:200px; text-align:center;" \|Text
		+	\|}

← Older revision		Revision as of 18:29, 4 February 2019
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	===About Resolving Forces===		===About Resolving Forces===
		+	: '''Resolving forces''' is used to find how much of a [[force]] acts in a certain direction. For example:
		+	:: It may be useful to know what 'component' of an [[object]]'s [[weight]] acts down the slope to tell if is likely to [[accelerate]] down the slope.
		+	:: A rope pulling at an [[angle]] above the [[horizontal]] has a [[vertical]] 'component' acting against [[weight]] and a [[horizontal]] 'component' acting to [[accelerate]] the [[object]] [[horizontal]]ly.
		+
		+	===Examples===