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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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Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

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