The weight vector of an object on an inclined plane can be split into its components parallel and perpendicular to the slope
A helicopter provides a lift of 250 kN when the blades are tilted at 15º from the vertical.Calculate the horizontal and vertical components of the lift force.
Step 1: Draw a vector triangle of the resolved forces
Step 2: Calculate the vertical component of the lift force
Vertical = 250 × cos(15) = 242 kN
Step 3: Calculate the horizontal component of the lift force
Horizontal = 250 × sin(15) = 64.7 kN
A person is exploring a new part of town, from their starting point they walk 100 m in the direction 30.0º South of West. They then walk 200 m in the direction 40.0º degrees South of East and finally they walk 150 m directly East.Calculate the magnitude of their displacement from their original position.
In order to calculate the answer, the vectors of displacement must be resolved into their x-components and y-components and then combined. In this case, this effectively means the x-direction is East-West and the y-direction is North-South
Step 1: Consider positive and negative directions for reference
Step 2: Resolve the first displacement (100 m magnitude) into its components
cos(30°) × 100 = 86.6 m
sin(30°) × 100 = 50.0 m
Step 3: Resolve the second displacement (200 m magnitude) into its components
cos(40°) × 200 = 153 m
sin(40°) × 200 = 129 m
Step 4: Resolve the third displacement (150 m magnitude) into its components
150 m
Step 5: Combine the horizontal (East-West) components
153 + 150 - 86.6 = 166 m
Step 6: Combine the vertical (North-South) components
50.0 + 129 = 170 m
Step 7: Using Pythagoras theorem to find the resultant hypotenuse vector
√(1662 + 1792) = 244 m
Three forces on an object in equilibrium form a closed vector triangle
A weight hangs in equilibrium from a cable at point X. The tensions in the cables are T1 and T2 as shown.Which diagram correctly represents the forces acting at point X?
If you're unsure as to which component of the force is cos θ or sin θ, just remember that the cos θ is always the adjacent side of the right-angled triangle AKA, making a 'cos sandwich'
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