What is the magnitude of an equal and opposite force F3, which balances the first two forces

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Creighton University

Tina W.

Physics 101 Mechanics

1 year ago

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Forces F, Fz, and Fz are all exerted on object; adding together to form a net force vector; Fnet = EF as shown in the graph to the right: However; only vectors Fi, Fz, and Fnet are known On a separate piece of graph paper; use the properties of vector addition to graphically determine the vector F3 eEiteEE 2 on object Lon_oblecth object A Brief Review of Trigonometry If you know the X- and y- components of a vector; A = (Ax; Ay), and you want t0 determine the magnitude (written as |A| or just A) and direction of a vector, you need to use trigonometry. The magnitude is the length of the vector arTow, (for example, the speed of a velocity vector) , and the direction is given by angle 0 relative to a chosen coordinate system. See figure on the right: Using trigonometry, we get: M| = VAr+ 4} 0 = 4z tan Ax If you are given the magnitude and direction of a vector; but want to express it in terms of its x and components, then you can use the following trigonometric properties: Ax = Acos 0 Ay = Asin 0 Make sure you are comfortable with vector trigonometry, since we will be using it a lot in the coming weeks

In each situation below, three forces are acting on a particle. Given two of the forces and the fact the particle is in equilibrium, find the third force.

\(\mathbf{a} = \quantity{-3\mathbf{i} + 2\mathbf{j}}{N}\)

\(\mathbf{b} = \quantity{4\mathbf{i} + \mathbf{j}}{N}\)

\(\mathbf{a}\) has magnitude \(\quantity{10}{N}\) and direction \(45^{\circ}\) anticlockwise from the positive \(x\) axis

\(\mathbf{b}= \quantity{6\mathbf{i} - 3\mathbf{j}}{N}\)

Is it possible to use a different approach for each situation?

In each case the particle is in equilibrium, which means the forces are balanced in all directions and the net force acting on the particle is zero. Below we discuss three possible approaches.

Triangle of forces

Since the particle is in equilibrium we can draw a triangle of forces and see approximately what the third force might be.

Why must the third force join up the other two forces to make a triangle?

Using the geometry of the triangle, we can find the magnitude of the third force with the cosine rule.

\[\begin{align*} \mathbf{|c|}^2 &= 20^2 + 35^2 - 2 \times 20 \times 35 \times \cos60 \\ \mathbf{|c|} &= 5\sqrt{37} \\ &\approx \quantity{30.4}{N} \end{align*}\]

To describe the third force fully we need the direction it is acting in. Many methods could be used, but we created two right angled triangles that share a side.

\[\begin{align*} 5\sqrt{37}\cos\theta &=20\cos 30 \\ \cos\theta &= \dfrac{2\sqrt{3}}{\sqrt{37}} \\ \theta &= 55.3^{\circ} \text{ (3 s.f.)} \end{align*}\]

We can draw the force on the original diagram.

How would we describe this vector in words?

We could have also used Lami’s Theorem, which states that if three forces, acting from a point, are in equilibrium, then the magnitude of each force is proportional to the sine of the angle between the other two forces.

\(\dfrac{\mathbf{|a|}}{\sin \alpha} = \dfrac{\mathbf{|b|}}{\sin \beta} = \dfrac{\mathbf{|c|}}{\sin \gamma}\)

Sum of forces

\(\mathbf{a} = \quantity{-3\mathbf{i} + 2\mathbf{j}}{N}\)

\(\mathbf{b} = \quantity{4\mathbf{i} + \mathbf{j}}{N}\)

It is always useful to draw a diagram. It can help to check whether answers seem sensible.

Where would you estimate the third force to be?

Since the particle is in equilibrium the sum of all the forces acting on it must be zero, i.e. \(\mathbf{a} + \mathbf{b} + \mathbf{c} = 0\). Therefore

\[-3\mathbf{i} + 2\mathbf{j} + 4\mathbf{i} + \mathbf{j} + \mathbf{c} = 0,\]

so the third force is \(\mathbf{c} =\quantity{-\mathbf{i} - 3\mathbf{j}}{N}.\)

What are the similarities between this approach and drawing a triangle of forces as used in the first situation?

Resolving forces

\(\mathbf{a}\) has magnitude \(\quantity{10}{N}\) and direction \(45^{\circ}\) anticlockwise from the positive \(x\) axis

\(\mathbf{b}= \quantity{6\mathbf{i} - 3\mathbf{j}}{N}\)

Resolving forces means we break forces down into perpendicular components (usually horizontal and vertical). Since it is in equilibrium we know that the sum of the horizontal forces must be zero, as must the vertical forces. If our third force is \(\mathbf{c} = p\mathbf{i}+q\mathbf{j}\) then we have the following diagram.

Resolving in the horizontal direction:	Resolving in the vertical direction:
\[6 + 10\cos45 + p = 0\] \[\implies p = -6-5\sqrt{2}\]	\[10\sin45 - 3 + q = 0\] \[\implies q = 3-5\sqrt{2}\]

So \(\mathbf{c} = \left(-6-5\sqrt{2}\right)\mathbf{i} + \left(3 - 5\sqrt{2}\right)\mathbf{j}.\)

What are the similarities and differences between these approaches?

Would each of these approaches work equally well for the different situations? If not, why not?

I will start by deconstructing the problem.

An object acted on by three forces moves with a constant velocity

If an object is moving at constant velocity, it means that it has an acceleration of 0. Since #F=ma#, this also means that there is no net force acting on the object.

#therefore F_"net" = 0 N /_ 0^@#

(remember that force not only has a magnitude (in Newtons) but also direction (in degrees). I'm setting the positive x-axis as 0 degrees for this problem, as it's very common.)

One force acting on the object is in the positive x direction and has a magnitude of 6.5 N; a second force has a magnitude of 4.4N and points in the negative y direction.

We are given two forces:
#F_1 = 6.5 N /_ 0^@#
#F_2 = 4.4N /_ 270^@#

Find the direction and magnitude of the third force acting on the object.

We know two of three forces, as well as the net force on the object. We can now calculate the third force.

We should start by adding #F_1# and #F_2# together to see what the current net force on the object is.

#F_1 + F_2 #

#magnitude = sqrt(6.5^2 + 4.4^2) = 7.85N#

(note that the two triangles form a right triangle, making this easy. If not, there would be more trigonometry here - calculating the horizontal and vertical magnitudes separately before similarly combining them. Look more into vector addition.)

#angle = tan^-1 (4.4/6.5) = -34.09^@ = 325.91^@#
#F_1 + F_2 = 7.85N angle 325.91^@#

Now that we know #F_1 + F_2 #, we can figure out an #F_3# to cancel it out. In our case, we simply need a force of equal magnitude in the opposite direction. The opposite direction of #325.91^@# is #325.91-180 = 145.91^@#.

#therefore F_3 = 7.85N angle 145.91^@ #

#square#