Simon's copy of Resolve forces on particle in equilibrium

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Simon Thomas 5 years, 11 months ago

Created this as a copy of Resolve forces on particle in equilibrium.

Name	Status	Author	Last Modified
Resolve forces on particle in equilibrium	Ready to use	Amy Chadwick	04/08/2016 11:56
Q1 Constant Acceleration	draft	Stephen Bowlzer	09/08/2016 13:08
Q2 Constant Acceleration	draft	Stephen Bowlzer	09/08/2016 14:57
John's copy of Resolve forces on particle in equilibrium	draft	John Bridges	19/09/2016 00:00
Simon's copy of Resolve forces on particle in equilibrium	draft	Simon Thomas	04/04/2019 15:26

Question statement

The diagram shows a particle in equilibrium under the forces shown.

You are told that $F_1 = \var{F1} \ \mathrm{N}$ and that the angles are $\theta_1 = \var{theta1}^{\circ}, \ \theta_2 = \var{theta2}$ and $\theta_3 = \var{theta3}^{\circ}$ .

By resolving forces, find the magnitude of the forces $F_2$ and $F_3$ , in Newtons, and input them to 3 d.p. below.

Give any introductory information the student needs.

			Name	Type	Generated Value

F1	number	2.5
theta1	integer	59
F2	number	1.4033223238
theta2	integer	70
F3	number	1.9502863319
theta3	integer	25
check	number	0
check2	number	0

Automatically regenerate variables?

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A random number between and (inclusive) with step size

Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `number`

2.5

→ Used by:

F2
F3
check
check2

Statement
Advice

Parts

Part a) Gap-fill
1. Gap Number entry
  Add
Part b) Gap-fill
1. Gap Number entry
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

$F_2 =$ Unnamed gap

Advice

As the particle is in equilibrium we equate the sum of the forces to zero.

Method 1

The quickest method to find $F_2$ is to resolve in the direction of $F_3$ . Then:

$F_1 \sin(\theta_1-\theta_3) = F_2 \sin(\theta_2+\theta_3)$

$F_2 = F_1 \frac{\sin(\theta_1-\theta_3)}{\sin(\theta_2+\theta_3)}$

$F_2 = \var{F1} \frac{\sin(\var{theta1}-\var{theta3})}{\sin(\var{theta2}+\var{theta3})}=\var{F1} \frac{\sin(\var{theta1-theta3})}{\sin(\var{theta2+theta3})}=\var{precround(F1*sin((pi/180)*(theta1-theta3))/sin((pi/180)*(theta2+theta3)),3)}$

Similarly, the quickest method to find $F_3$ is to resolve in the direction of $F_2$ . Then:

$F_1 \sin(\theta_1+\theta_2) = F_3 \sin(\theta_2+\theta_3)$

$F_3 = F_1 \frac{\sin(\theta_1+\theta_2)}{\sin(\theta_2+\theta_3)}$

$F_3 = \var{F1} \frac{\sin(\var{theta1}+\var{theta2})}{\sin(\var{theta2}+\var{theta3})}=\var{F1} \frac{\sin(\var{theta1+theta2})}{\sin(\var{theta2+theta3})}=\var{precround(F1*sin((pi/180)*(theta1+theta2))/sin((pi/180)*(theta2+theta3)),3)}$

Method 2

Alternatively, we can find $F_2$ and $F_3$ by resolving horizontally and vertically as follows:

Resolving horizontally we have

$\begin{align} F_1 \cos \theta_1 + F_2 \cos \theta_2 - F_3 \cos \theta_3 & = 0, \\ \var{F1} \cos \var{theta1}^{\circ} + F_2 \cos \var{theta2}^{\circ} - F_3 \cos \var{theta3}^{\circ} & = 0, \\ F_2 \cos \var{theta2}^{\circ} & = F_3 \cos \var{theta3}^{\circ} - \var{F1} \cos \var{theta1}^{\circ}, \\ F_2 & = \frac{F_3 \cos \var{theta3}^{\circ} - \var{F1} \cos \var{theta1}^{\circ}}{ \cos \var{theta2}^{\circ}}. \end{align}$

Resolving vertically, we have

$\begin{align} F_1 \sin \theta_1 - F_2 \sin \theta_2 - F_3 \sin \theta_3 & = 0, \\ \var{F1} \sin \var{theta1}^{\circ} - F_2 \sin \var{theta2}^{\circ} - F_3 \sin \var{theta3}^{\circ} & = 0, \\ F_3 \sin \var{theta3}^{\circ}& = \var{F1} \sin \var{theta1}^{\circ} - F_2 \sin \var{theta2}^{\circ}. \end{align}$

Substituting our value for $F_2$ into the formula $F_3 \sin \var{theta3}^{\circ}$ and rearranging gives

$\begin{align} F_3 \sin \var{theta3}^{\circ} & = \var{F1} \sin \var{theta1}^{\circ} - F_2 \sin \var{theta2}^{\circ} \\ F_3 \sin \var{theta3}^{\circ} & = \var{F1} \sin \var{theta1}^{\circ} - \left( \frac{F_3 \cos \var{theta3}^{\circ} - \var{F1} \cos \var{theta1}^{\circ}}{\cos \var{theta2}^{\circ}} \right) \sin \var{theta2}^{\circ}, \\ F_3 \sin \var{theta3}^{\circ} & = \var{F1} \sin \var{theta1}^{\circ} - F_3 \frac{ \cos \var{theta3}^{\circ}}{ \cos \var{theta2}^{\circ}} \sin \var{theta2}^{\circ} + \var{F1} \frac{ \cos \var{theta1}^{\circ}}{ \cos \var{theta2}^{\circ}} \sin \var{theta2}^{\circ}, \\ F_3 \left( \sin \var{theta3}^{\circ} + \frac{ \cos \var{theta3}^{\circ}}{ \cos \var{theta2}^{\circ}} \sin \var{theta2}^{\circ} \right) & = \var{F1} \sin \var{theta1}^{\circ} + \var{F1} \frac{ \cos \var{theta1}^{\circ}}{ \cos \var{theta2}^{\circ}} \sin \var{theta2}^{\circ} \\ F_3 & = \frac{\var{F1} \sin \var{theta1}^{\circ} + \var{F1} \frac{ \cos \var{theta1}^{\circ}}{ \cos \var{theta2}^{\circ}} \sin \var{theta2}^{\circ}}{ \sin \var{theta3}^{\circ} + \frac{ \cos \var{theta3}^{\circ}}{ \cos \var{theta2}^{\circ}} \sin \var{theta2}^{\circ}}. \end{align}$

Therefore, to 3 s.f. we have $F_3 = \var{precround(F3,3)} \mathrm{N}.$

We can then use our $F_3$ value in our equation for $F_2$

$\begin{align} F_2 & = \frac{F_3 \cos \var{theta3}^{\circ} - \var{F1} \cos \var{theta1}^{\circ}}{\cos \var{theta2}^{\circ}}, \\ & = \frac{\var{precround(F3,3)} \cos \var{theta3}^{\circ} - \var{F1} \cos \var{theta1}^{\circ}}{\cos \var{theta2}^{\circ}}, \\ & = \var{precround(F2,3)} \mathrm{N}. \end{align}$

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This question is used in the following exam:

Statics by Simon Thomas in Mechanics.

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Simon's copy of Resolve forces on particle in equilibrium

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Simon Thomas 5 years, 11 months ago

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Simon's copy of Resolve forces on particle in equilibrium

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Taxonomy: mathcentre

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Simon Thomas 5 years, 11 months ago

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Generated value: number

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Generated value: `number`