Particle in vertical equilibrium, accelerating horizontally

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Christian Lawson-Perfect	said	Ready to use	8 years, 7 months ago
Amy Chadwick	said	Needs to be tested	8 years, 9 months ago

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Christian Lawson-Perfect 8 years, 7 months ago

Gave some feedback: Ready to use

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Christian Lawson-Perfect 8 years, 7 months ago

Saved a checkpoint:

Clarified the wording in the statement a bit, and it seems that the other things have been fixed.

Restore this checkpoint

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Christian Lawson-Perfect 8 years, 8 months ago

Gave some feedback: Has some problems

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Christian Lawson-Perfect commented 8 years, 9 months ago

Oh, and again you need to say in the prompt what precision you want, when you set a precision restriction in the marking.

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Christian Lawson-Perfect commented 8 years, 9 months ago

In GeoGebra, hide the points that don't need to be drawn, such as the ones on either end of the acceleration vector.
a)
Need to specify units in "the mass of the particle $K = ...$ ". Stick an $\mathrm{N}$ after the input to show that the answer is in Newtons. The input line should go

$C =$ [[0]] $\mathrm{N}$
b)
Use \simplify - I got $1gN$ when FD = 1. You need to say which direction to resolve the force in, or ask to find the magnitude of the force.
c)
Decelerating in which direction? You've set up a 2D problem sort of needlessly, because the components are completely independent.

Why change the mass of the particle? Say you're considering a different particle. How is this part different to part a?
d)
Again, this is the same as part b but with different numbers - maybe split this into two separate questions. And again you need to use \simplify when giving the force $B$ .

The prompt should say "now there is a net force of ... acting downwards". But how do you know the net force? You'd work it out from observing the acceleration. This question could go as follows:

Set up: the particle has mass $K\, \mathrm{kg}$ , and is accelerating at $a\, \mathrm{ms^{-2}}$ vertically downwards.

First part: What is the total force acting on the particle?

Second part: a downwards force of $D \, \mathrm{N}$ and a downwards force $B$ are being applied to the particle. What is the magnitude of $B$ , in Newtons?

You could have one question considering vertical acceleration, and another for horizontal.

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Christian Lawson-Perfect commented 8 years, 9 months ago

Use \mathrm on units so you don't get italics, e.g. $5\mathrm{ms}^{-1}$

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Amy Chadwick 8 years, 9 months ago

Gave some feedback: Needs to be tested

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Amy Chadwick 8 years, 9 months ago

Created this.

Name	Status	Author	Last Modified
Particle in vertical equilibrium, accelerating horizontally	Ready to use	Amy Chadwick	24/08/2016 10:50
Particle in vertical equilibrium, accelerating horizontally	Needs to be tested	Chris Graham	08/12/2016 11:23
Simon's copy of Particle in vertical equilibrium, accelerating horizontally	draft	Simon Thomas	29/04/2019 09:19

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Question statement

Consider the following diagram of a particle of mass $K \ \mathrm{kg}$ , moving left-to-right. The forces $A, B, C$ and $D$ are acting upon it, producing a horizontal acceleration of $a \, \mathrm{ms^{-2}}$

The acceleration due to gravity is $9.8 \, \mathrm{ms}^{-2}$ . Answer all the following questions to 3 decimal places.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	number	6.5
K	integer	7
FA	number	0.5
FD	integer	7
a2	number	0.25
K2	integer	4
FA2	number	78.5
downforce	integer	8
FD2	integer	4

Automatically regenerate variables?

Name

Data type

Value

A random number between and (inclusive) with step size

Description

acceleration

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

Can an exam override the value of this variable?

Generated value: `number`

6.5

→ Used by:

Advice
"Part a)" - prompt
"Part a)" → "Unnamed gap" - Minimum accepted value
"Part a)" → "Unnamed gap" - Maximum accepted value

Parts

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

Gap-fill

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

Suppose that the acceleration $a = \var{a}$ and the mass of the particle $K = \var{K} \ \mathrm{kg}$ . Find the force $C \ \mathrm{N}$ if $A = \var{FA}\ \mathrm{N}$ .

$C =$ Unnamed gap $\mathrm{N}$

Advice

a)

To find the resultant force in a certain direction when there is more than one force acting on an object you can resolve the forces. You resolve in the direction of acceleration. Using the letter $R$ , with an arrow will indicate the direction the forces are being resolved in, for example $R(\rightarrow)$ .

In this case $R(\rightarrow)$ means apply $F=ma$ in the positive direction. So we need to resolve the equation

$\begin{align}F & = ma \\ C - \var{FA} & = \var{K} \times \var{a} \\ C & = \var{K*a} + \var{FA} \\ & = \var{K*a + FA}. \end{align}$

Here the forces, $F=C - \var{FA}$ because $\var{FA}$ is acting in the opposite direction to $C$ . The mass is $m=\var{K}$ and the acceleration is $a=\var{a}$ . Therefore $C = \var{K*a+FA}\ \mathrm{N}$ .

b)

In this case we resolve in the direction perpendicular to acceleration so we use $R(\uparrow)$ which means apply $F=ma$ in the vertical direction, where acceleration, $a=0$ because there is no vertical acceration.

$\begin{align} F & = ma \\ B - \var{FD}g & = \var{K} \times 0 \\ B & = 0 + \left(\var{FD} \times 9.8\right) \\ & = \var{FD*9.8} \end{align}$
Here the forces,

$F=B - \var{FD}g$ because

$\var{FD}g$ is acting downwards which is the opposite direction to

$B$ . Therefore

$B = \var{FD*9.8} \ \mathrm{N}$ .

c)

It is easier to take the positive direction as the direction of the acceleration. Therefore we use $R(\leftarrow)$ because we are decelerating.

$\begin{align} F & = ma \\ \var{FA2} - C & = \var{K2} \times \var{a2} \\ C & = \var{FA2} - \left( \var{K2} \times \var{a2} \right) \\ & = \var{FA2 - K2*a2}. \end{align}$

Here the forces, $F = \var{FA2} - C$ because $\var{FA2}$ is in the direction of acceleration and $C$ is acting in the opposite direction. The mass is $m = \var{K2}$ and $a=\var{a2}$ . Therefore $C = \var{FA2 - K2*a2} \ \mathrm{N}$ .

Give a worked solution to the whole question.

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This question is used in the following exam:

Dynamics - Forces by Amy Chadwick in Maths Support Wiki - Mechanics.

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Particle in vertical equilibrium, accelerating horizontally

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Checkpoint description

Christian Lawson-Perfect 8 years, 7 months ago

Christian Lawson-Perfect 8 years, 7 months ago

Christian Lawson-Perfect 8 years, 8 months ago

Christian Lawson-Perfect commented 8 years, 9 months ago

Christian Lawson-Perfect commented 8 years, 9 months ago

Christian Lawson-Perfect commented 8 years, 9 months ago

Amy Chadwick 8 years, 9 months ago

Amy Chadwick 8 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: `number`

Parts

Gap-fill

a)

b)

c)

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Error

Particle in vertical equilibrium, accelerating horizontally

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 8 years, 7 months ago

Christian Lawson-Perfect 8 years, 7 months ago

Christian Lawson-Perfect 8 years, 8 months ago

Christian Lawson-Perfect commented 8 years, 9 months ago

Christian Lawson-Perfect commented 8 years, 9 months ago

Christian Lawson-Perfect commented 8 years, 9 months ago

Amy Chadwick 8 years, 9 months ago

Amy Chadwick 8 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

a)

b)

c)

Generated value: `number`