find by drawing and by caculation the resultant of two vectors 3 units and 4 units inclined to each other at 90 degrees
1 answer:
The resultant of the two vectors is 5 units (see attachment)
Explanation:
When we have to calculate the resultant of two vectors, we can do the following:
- Graphically, we have to draw the two vectors, connected head-to-tail (this means, the tail of the second vector starts from the head of the first vector). The resultant vector is found by connecting the tail of the first vector to the head of the second vector. In this example, we called A the first vector of 3 units and B the second vector of 4 units: the rresultant vector is shown in the picture as vector R.
- Numerically, we start by noticing that the two vectors are at 90 degrees to each other. Therefore, they form the sides of a right triangle, of which the resultant is the hypothenuse.
Therefore, we can find the resultant by using Pythagorean's theorem. Using
A = 3
B = 4
We find:
So, the resultant is 5 units.
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Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)
• The net force in the parallel direction is
∑ <em>F</em> (para) = -<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
• The net force in the perpendicular direction is
∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0
Solving the second equation for <em>n</em> gives
<em>n</em> = <em>mg</em> cos(21°)
<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)
<em>n</em> ≈ 1.83 N
Then the magnitude of friction is
<em>f</em> = <em>µn</em>
<em>f</em> = 0.25 (1.83 N)
<em>f</em> ≈ 0.457 N
Solve for the acceleration <em>a</em> :
-<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
<em>a</em> = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)
<em>a</em> ≈ -5.80 m/s²
so the block is decelerating with magnitude
<em>a</em> = 5.80 m/s²
down the ramp.
By Newton's second law, the apparent weight of rider coming at rest is <u>3548 N</u>.
According to formula, v = u + at
a = 39.2 m/
Force, F = ma
F = m(a+g)
F = apparent weight
m = mass, given = 65 kg
a = acceleration, given = 39.2 m/
g = 9.8 m/
Put these values in formula, F = m(a+g)
F = 65(39.2+9.8)
F = 3548 N
Therefore, the apparent weight of rider coming at rest is <u>3548 N</u>.
Learn more about apparent weight here:- brainly.com/question/24897276
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