Scalars & Vectors
The difference between 'how much' and 'which way' -- resolving and combining forces like a physicist.
Spec Points Covered
- I can define what scalar and vector quantities are and give examples of each.
- I can add two vectors using the triangle of vectors method.
- I can add two vectors using the parallelogram of vectors method.
- I can resolve a single vector into two perpendicular components using trigonometry.
- I can calculate the magnitude and direction of a resultant from two perpendicular components.
- I can draw and interpret a free body diagramA diagram showing all the forces acting on a single object, drawn as arrows from the centre of the object. showing all forces acting on an object.
- I can explain the condition for equilibriumAn object is in equilibrium when the resultant force on it is zero. The object is either stationary or moving at constant velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. using a closed triangle of forces.
- I can use the conditions for equilibriumAn object is in equilibrium when the resultant force on it is zero. The object is either stationary or moving at constant velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. to solve problems involving three coplanar forces.
- I can resolve forces on an inclined plane into components parallel and perpendicular to the slope.
- I can determine a resultant vector by scale drawing and by calculation.
Notes
01
Scalar quantity
Scalar quantity
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02
Vectors cannot be added like ordinary numbers unless they act along the same line
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03
In the parallelogram of vectors method
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Any single vector can be split into two perpendicular components
$F_x = F\cos\theta$
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If you know the horizontal and vertical components
$F = \sqrt{F_x^2 + F_y^2}$
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Equilibrium
Equilibrium
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07
Free body diagram
Free body diagram
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08
On a slope at angle theta to the horizontal
$F_{\parallel} = mg\sin\theta$
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09
A rope pulls a box along a floor with a force of 120 N at 25 degrees above the horizontal
$F_x = F\cos\theta = 120 \times \cos(25^\circ) = 120 \times 0.906 = 108.8 \text{ N}$
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10
A lamp of weight 20 N hangs from two strings
$T_A \cos(30^\circ) = T_B \cos(50^\circ)$
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On Data Sheet
Not on Data Sheet
Horizontal component of a vector
$$F_x = F\cos\theta$$
- Where:
- $F_x$ = same units as F
- $F$ = magnitude of the vector
- $\theta$ = angle between the vector and the horizontal (degrees or radians)
Use when the angle is measured from the horizontal. The adjacent component uses cos. If the angle is from the vertical, cos gives the vertical component instead.
Vertical component of a vector
$$F_y = F\sin\theta$$
- Where:
- $F_y$ = same units as F
- $F$ = magnitude of the vector
- $\theta$ = angle between the vector and the horizontal (degrees or radians)
Use when the angle is measured from the horizontal. The opposite component uses sin. Always draw the right-angled triangle to confirm which trig function to use.
Resultant of two perpendicular vectors
$$F = \sqrt{F_x^2 + F_y^2}$$
- Where:
- $F$ = resultant magnitude (same units as components)
- $F_x$ = horizontal component
- $F_y$ = vertical component
Pythagoras' theorem applied to perpendicular components. Only valid when the two components are at 90 degrees to each other.
Angle of resultant vector
$$\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)$$
- Where:
- $\theta$ = degrees or radians
- $F_y$ = component opposite to the angle
- $F_x$ = component adjacent to the angle
Gives the angle measured from the horizontal (if F_x is horizontal and F_y is vertical). Always state the reference direction in your answer.
Weight component parallel to a slope
$$F_{\parallel} = mg\sin\theta$$
- Where:
- $F_{\parallel}$ = N
- $m$ = kg
- $g$ = N kg^-1 (or m s^-2)
- $\theta$ = angle of slope to horizontal
The component of weight that pulls the object down the slope. As the slope gets steeper (theta increases), this component increases. At theta = 90 degrees (vertical), it equals mg.
Weight component perpendicular to a slope
$$F_{\perp} = mg\cos\theta$$
- Where:
- $F_{\perp}$ = N
- $m$ = kg
- $g$ = N kg^-1 (or m s^-2)
- $\theta$ = angle of slope to horizontal
The component of weight pushing the object into the surface. Equals the normal contact force if there is no other perpendicular force. At theta = 0 (flat), this equals mg.