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Usimov [2.4K]
3 years ago
11

Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 58.0 ° counterclockwise from the + x ‑axi

s, whereas the second force has x ‑ and y ‑components ( − 19.0 N , 16.5 N ) . Express the magnitude and direction of the net force.
Physics
1 answer:
Ira Lisetskai [31]3 years ago
7 0
<h2>Answer:</h2>

magnitude of the net force is 32.49N

and the direction of the net force is 107.91°

<h2>Explanation:</h2>

Let the first force be F₁

Let the second force be F₂

Let the net force be F

<em>From the question;</em>

(a) F₁ has 17.0N in magnitude and has direction θ = 58.0° counterclockwise from +x axis. i.e

|F₁| = 17.0N

θ = 58.0°

<em>Re-write this force in vector form as follows;</em>

F₁ = |F₁| cos θ i + |F₁| sin θ j

F₁ = [17.0 cos 58.0°] i + [17.0 sin 58.0] j

F₁ = [17.0 x 0.5299] i + [17.0 x 0.8480] j

F₁ = [9.0083] i + [14.416] j

<em>Put the unit as follows;</em>

F₁ = [9.0083 N] i + [14.416 N] j

(b) F₂ is already written in vector form but can be re-written as follows;

F₂ = [-19.0N ] i + [16.5N] j

(c) The net force (F) is the algebraic sum of the two forces F₁ and F₂ as follows;

F = F₁ + F₂

<em>Substitute the values of F₁ and F₂ into the equation above;</em>

F =  [9.0083 N] i + [14.416 N] j + [-19.0N ] i + [16.5N] j

<em>Now, collect like terms;</em>

F =  [9.0083 N] i + [-19.0 N] i + [14.416N ] j + [16.5N] j

F = -9.9917 N i  + 30.916 N j                     ----------------(a)

===============================================================

(a) Now let's calculate the magnitude, |F|, of the net force as follows;

|F| = \sqrt{(-9.9917)^2 + (30.916)^2}

|F| = \sqrt{(99.83) + (955.80)}

|F| = \sqrt{1055.63}

|F| = 32.49N

Therefore, the magnitude of the net force is 32.49N

(b) The direction, θ, is calculated as follows;

tan θ = (y component / x component)

tan θ = (30.916 / -9.9917)

tan θ = (-3.094)

θ = tan ⁻¹ (-3.094)

θ = -72.09°

The negative sign shows that it is measured with respect to the -x axis.

From equation (a), it is shown that the net force is located in the -i and +j directions measured counterclockwise from the +x axis. Therefore, to get the true angle of the net force relative to the positive x axis and counterclockwise, we add 180° to the angle calculated above. i.e

true angle = 180° + (-72.09°)

true angle = 107.91°

Therefore, the direction of the net force is 107.91° counterclockwise from the +x axis.

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