A child pulls a sled through the snow with a force of 50 Newtons exerted at an angle of 30 degrees above the horizontal. Find the horizontal and vertical components of this force.
F=50cos(30)i+50cos(60)j
And is this Calculus 3 problem?
A child pulls a sled through the snow with a force of 50 Newtons exerted at an angle of 30 degrees above the horizontal. Find the horizontal and vertical components of this force.
F=50cos(30)i+50cos(60)j
And is this Calculus 3 problem?
This explanation from Physics/Geometry
60^{o} 

 F_{y} (the vert. comp.)
↑
← 30^{o} 
F_{x} (the horizontal componenet)
F = √(F_{x}^{2} + F_{y}^{2})
F_{y} = 50 · cos 60^{o} = 50 · (1/2) = 25 N
F_{x} = 50 · cos 30^{o} = 50 · (√3)/2 = 25√3 N
I see, that vector sign did not appear in my comment above, so the vector equation is
→ → →
F = 50 cos 30^{o} i + 50 cos 60^{o}
j
This force vector has a magnitude of 50 N at 30 degrees to the Horizontal. This vector has two components, one horizontal and one vertical. The horizontal component of the force vector is the vector projected on the horizontal x axis and the vertical component is the projection of the vector on the vertical y axis. The 50 N force is the resultant of the two components. therefore in a right angle triangle
The horizontal component can be found from cos 30 = x / 50, x = horizontal component of the force,
x= 50 cos 30 = 50 × (√3/2) = 43.3 N
To solve for the Y or vertical component of the force, in the right angle triangle the angle opposite to tthe 30 degree angle is 60 since the sum of all angles must be 180 degrees. Therefore either the cosine of the 60 degree angle or the sin of the 30 degree angle multiplied by the resultant force or the hypotenous of the right anle triangle formed gives the vertical component.
y = 50 cos 60 = 50 sin 30 = 50 × (1/2) = 25 N
The solution can be checked by finding R, it must be equal to 50 lb
In this right angle triangle and applying the phythogoren theorem R^{2} = x^{2} + y^{2} , R = √ x^{2} + y^{2}
R = √ (25 N)^{2}+(43.3 N)^{2} = 50 N and the solution checks
therefore the vector F can be written as F = 43.3 N i + 25 N j
Suppose you have a x,y plane(graph) so the horizontal is the xaxis while the vertical is the yaxis.
Let us represent the force of 50N as “vector”
If we plot a straight line with a measure of 50 (to represent vector) at an angle of 30degrees from the origin(starting point), we see that the line gradually rises above the horizontal and eventually ends when it reaches a measure of 50.
Now if we draw a vertical straight line from the end of the vector, straight down till we hit the horizontal(xaxis), this vertical line becomes the vertical component. Now, if we draw a straight horizontal line from the origin, to the point where the vertical
component hits the xaxis and stop. This new line is the horizontal component.
NOTE:
• We now have a right angle tringle with a hypotenuse and two legs. Hence we can use trig (SOH, CAH, TOA should work for this.)
• Notice that this triangle falls in the first quadrant so all components have to be positive.
Using the SOH CAH TOA: SOH implies
Sine=opposite(O)/hypotenuse (H)
Cosine =adjacent(A)/hypotenuse (H)
Tangent= opposite(O)/adjacent(A)
Here we have a hypotenuse (50), and an angle(30) so we have to find the opposite(O) which is the vertical component and the adjacent (A) is the horizontal component.
So sin30=O/H → sin30=O/50 → (multiply both sides by 50) → O=(50)*(sin30) → O=25N
And cos30=A/H → cos30=A/50 →(multiply both sides by 50) → A=(50)*(cos30) → A=43.3N
So the Vertical component is 25N and the horizontal component is 43.3N.
NOTICE: in the end, notice that we were just looking for the rise(vertical component) and the run(horizontal component)
This could be a physics, geometry of trigonometry question.
Comments
It's Physics with elements of Linear Algebra. Mathematics was always Physics's tool, and was separated from Physics, maybe, 35 years ago. At that time applied mathematics became a mathematical science.....
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F = 50 · cos 30^{o } · i + 50 · cos 60^{o} · j
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i and j are unit vectors.
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