Answers don't match! Mechanics 1 doubt

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Reshyna
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I did all the sections but when it came to c) part the answers didn't match.
My working:
R(upwards) R-10g-Tsin 20o=0
R= T sin 20o+10g

FMAX= 0.6(T sin 20o+10g)

R(sidewise) Tcos 20o - FMAX=2

Tcos 20o - 0.6(T sin 20o+10g) =2
Tcos 20o-0.6T sin 20o = 2+6g
T(cos 20o-0.6 sin 20o)=2+6g
T=(2+6g)/0.734(to 3s.f)
T=82.83N

but the answer given in the book is 53N
Does anyone have any ideas?
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K-Man_PhysCheM
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(Original post by Rashina)
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I did all the sections but when it came to c) part the answers didn't match.
My working:
R(upwards) R-10g-Tsin 20o=0
R= T sin 20o+10g

FMAX= 0.6(T sin 20o+10g)

R(sidewise) Tcos 20o - FMAX=2

Tcos 20o - 0.6(T sin 20o+10g) =2
Tcos 20o-0.6T sin 20o = 2+6g
T(cos 20o-0.6 sin 20o)=2+6g
T=(2+6g)/0.734(to 3s.f)
T=82.83N

but the answer given in the book is 53N
Does anyone have any ideas?
The box does not move up or down, so:

Sum of Up forces = Sum of down forces
or:
Normal + horizontal component of tension = Weight
Normal + T \sin 20^{\circ} = W
Hence:
Normal = W - T \sin 20^{\circ}

Try again with that value for the normal, and you should get the textbook's answer (I have worked it through).
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Reshyna
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(Original post by K-Man_PhysCheM)
The box does not move up or down, so:

Sum of Up forces = Sum of down forces
or:
Normal + horizontal component of tension = Weight
Normal + T \sin 20^{\circ} = W
Hence:
Normal = W - T \sin 20^{\circ}

Try again with that value for the normal, and you should get the textbook's answer (I have worked it through).
I got the answer. Thanks
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K-Man_PhysCheM
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(Original post by Rashina)
I got the answer. Thanks
Glad I could help
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Reshyna
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(Original post by K-Man_PhysCheM)
Glad I could help
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have any idea? This is the last one for today.
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K-Man_PhysCheM
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(Original post by Rashina)
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have any idea? This is the last one for today.
Here is a terrible diagram:
Name:  forcediagram.png
Views: 85
Size:  5.0 KB
Find the normal reaction force, hence find the friction force, then the resultant force for the motion, and then use F = ma.
Spoiler:
Show


R = W \cos \theta

Fr = \frac{1}{2} W \cos \theta = \dfrac{\cos (\tan^{-1} \frac{3}{4})}{2} mg = \frac{0.8}{2}mg = \frac{2}{5}mg

F = \dfrac{3mg}{2} = \frac{3}{2} mg

Resultant Force = Applied Force - Friction

F(res) =  \frac{3}{2} mg - \frac{2}{5} mg = 1.1mg

a = \dfrac{F(res)}{m} = \dfrac{1.1mg}{m} = 1.1g

a = 1.1 \times 9.8 = 10.78 = 10.8 ms^{-2}

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Reshyna
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(Original post by K-Man_PhysCheM)
Here is a terrible diagram:
Name:  forcediagram.png
Views: 85
Size:  5.0 KB
Find the normal reaction force, hence find the friction force, then the resultant force for the motion, and then use F = ma.
Spoiler:
Show



R = W \cos \theta

Fr = \frac{1}{2} W \cos \theta = \dfrac{\cos (\tan^{-1} \frac{3}{4})}{2} mg = \frac{0.8}{2}mg = \frac{2}{5}mg

F = \dfrac{3mg}{2} = \frac{3}{2} mg

Resultant Force = Applied Force - Friction

F(res) =  \frac{3}{2} mg - \frac{2}{5} mg = 1.1mg

a = \dfrac{F(res)}{m} = \dfrac{1.1mg}{m} = 1.1g

a = 1.1 \times 9.8 = 10.78 = 10.8 ms^{-2}


I can understand the diagram and the answer. Thanks.
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Reshyna
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(Original post by K-Man_PhysCheM)
Here is a terrible diagram:
Name:  forcediagram.png
Views: 85
Size:  5.0 KB
Find the normal reaction force, hence find the friction force, then the resultant force for the motion, and then use F = ma.
Spoiler:
Show



R = W \cos \theta

Fr = \frac{1}{2} W \cos \theta = \dfrac{\cos (\tan^{-1} \frac{3}{4})}{2} mg = \frac{0.8}{2}mg = \frac{2}{5}mg

F = \dfrac{3mg}{2} = \frac{3}{2} mg

Resultant Force = Applied Force - Friction

F(res) =  \frac{3}{2} mg - \frac{2}{5} mg = 1.1mg

a = \dfrac{F(res)}{m} = \dfrac{1.1mg}{m} = 1.1g

a = 1.1 \times 9.8 = 10.78 = 10.8 ms^{-2}


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I did this question but I could only get the answer in the book, when I divide the answer I worked out by two. Can you check it?
My working:
For A: R(upwards) T-1.2g=1.2a
For B: R(downwards) 0.9g+0.7g-T= 0.7a+0.9a
Adding both equations:
0.9g+0.7g-T+ T-1.2g = 1.2a+0.7a+0.9a
0.8g=2.8a
a=0.8g/2.8
a=2.8ms-1

But book's answer is 1.4ms-1.
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K-Man_PhysCheM
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(Original post by Rashina)
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I did this question but I could only get the answer in the book, when I divide the answer I worked out by two. Can you check it?
My working:
For A: R(upwards) T-1.2g=1.2a
For B: R(downwards) 0.9g+0.7g-T= 0.7a+0.9a
Adding both equations:
0.9g+0.7g-T+ T-1.2g = 1.2a+0.7a+0.9a
0.8g=2.8a
a=0.8g/2.8
a=2.8ms-1

But book's answer is 1.4ms-1.
Check your addition on the 4th line:

0.9g + 0.7g - 1.2g +T - T = 0.4g, NOT 0.8g
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Reshyna
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(Original post by K-Man_PhysCheM)
Check your addition on the 4th line:

0.9g + 0.7g - 1.2g +T - T = 0.4g, NOT 0.8g
Sorry for that. My bad. Thanks
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Reshyna
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(Original post by K-Man_PhysCheM)
Check your addition on the 4th line:

0.9g + 0.7g - 1.2g +T - T = 0.4g, NOT 0.8g
no doubts.just ignore the image
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RDKGames
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(Original post by K-Man_PhysCheM)
Here is a terrible diagram:
Name:  forcediagram.png
Views: 85
Size:  5.0 KB
Find the normal reaction force, hence find the friction force, then the resultant force for the motion, and then use F = ma.
Spoiler:
Show



R = W \cos \theta

Fr = \frac{1}{2} W \cos \theta = \dfrac{\cos (\tan^{-1} \frac{3}{4})}{2} mg = \frac{0.8}{2}mg = \frac{2}{5}mg

F = \dfrac{3mg}{2} = \frac{3}{2} mg

Resultant Force = Applied Force - Friction

F(res) =  \frac{3}{2} mg - \frac{2}{5} mg = 1.1mg

a = \dfrac{F(res)}{m} = \dfrac{1.1mg}{m} = 1.1g

a = 1.1 \times 9.8 = 10.78 = 10.8 ms^{-2}


Incorrect working.

Net force = F - friction force - gravity force parallel to the plane

So you missed out the gravity force.

I don't have time to check others so make sure you don't stop trip up with those when helping others.
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K-Man_PhysCheM
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(Original post by RDKGames)
Incorrect working.

Net force = F - friction force - gravity force parallel to the plane

So you missed out the gravity force.

I don't have time to check others so make sure you don't stop trip up with those when helping others.
Ah yes, my bad. Thank you for checking!
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Reshyna
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(Original post by K-Man_PhysCheM)
Ah yes, my bad. Thank you for checking!
Help me in this question,please.
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the c) part
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Reshyna
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(Original post by K-Man_PhysCheM)
Check your addition on the 4th line:

0.9g + 0.7g - 1.2g +T - T = 0.4g, NOT 0.8g
Can you help me with this question, please?
http://www.thestudentroom.co.uk/show...2#post68458290
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