# Answers don't match! Mechanics 1 doubt

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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 20

^{o}=0

R= T sin 20

^{o}+10g

F

_{MAX}= 0.6(T sin 20

^{o}+10g)

R(sidewise) Tcos 20

^{o }- F

_{MAX}=2

Tcos 20

^{o }- 0.6(T sin 20

^{o}+10g) =2

Tcos 20

^{o}-0.6T sin 20

^{o }= 2+6g

T(cos 20

^{o}-0.6 sin 20

^{o})=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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#2

(Original post by

I did all the sections but when it came to c) part the answers didn't match.

R(upwards) R-10g-Tsin 20

R= T sin 20

F

R(sidewise) Tcos 20

Tcos 20

Tcos 20

T(cos 20

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?

**Rashina**)I did all the sections but when it came to c) part the answers didn't match.

**My working:**R(upwards) R-10g-Tsin 20

^{o}=0R= T sin 20

^{o}+10gF

_{MAX}= 0.6(T sin 20^{o}+10g)R(sidewise) Tcos 20

^{o }- F_{MAX}=2Tcos 20

^{o }- 0.6(T sin 20^{o}+10g) =2Tcos 20

^{o}-0.6T sin 20^{o }= 2+6gT(cos 20

^{o}-0.6 sin 20^{o})=2+6gT=(2+6g)/0.734(to 3s.f)

T=82.83N

but the answer given in the book is 53N

Does anyone have any ideas?

Sum of Up forces = Sum of down forces

or:

Normal + horizontal component of tension = Weight

Normal + = W

Hence:

Normal =

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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(Original post by

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 + = W

Hence:

Normal =

Try again with that value for the normal, and you should get the textbook's answer (I have worked it through).

**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 + = W

Hence:

Normal =

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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(Original post by

Glad I could help

**K-Man_PhysCheM**)Glad I could help

have any idea? This is the last one for today.

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#6

Find the normal reaction force, hence find the friction force, then the resultant force for the motion, and then use F = ma.

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(Original post by

Here is a terrible diagram:

Find the normal reaction force, hence find the friction force, then the resultant force for the motion, and then use F = ma.

**K-Man_PhysCheM**)Here is a terrible diagram:

Find the normal reaction force, hence find the friction force, then the resultant force for the motion, and then use F = ma.

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**K-Man_PhysCheM**)

Here is a terrible diagram:

Find the normal reaction force, hence find the friction force, then the resultant force for the motion, and then use F = ma.

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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#9

(Original post by

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?

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

But book's answer is 1.4ms

**Rashina**)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}.0.9g + 0.7g - 1.2g +T - T =

**0.4g**, NOT 0.8g

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(Original post by

Check your addition on the 4th line:

0.9g + 0.7g - 1.2g +T - T =

**K-Man_PhysCheM**)Check your addition on the 4th line:

0.9g + 0.7g - 1.2g +T - T =

**0.4g**, NOT 0.8g
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**K-Man_PhysCheM**)

Check your addition on the 4th line:

0.9g + 0.7g - 1.2g +T - T =

**0.4g**, NOT 0.8g

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#12

**K-Man_PhysCheM**)

Here is a terrible diagram:

Find the normal reaction force, hence find the friction force, then the resultant force for the motion, and then use F = ma.

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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#13

(Original post by

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.

**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.

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(Original post by

Ah yes, my bad. Thank you for checking!

**K-Man_PhysCheM**)Ah yes, my bad. Thank you for checking!

the c) part

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**K-Man_PhysCheM**)

Check your addition on the 4th line:

0.9g + 0.7g - 1.2g +T - T =

**0.4g**, NOT 0.8g

http://www.thestudentroom.co.uk/show...2#post68458290

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