# Answers don't match! Mechanics 1 doubt

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#1

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?
0
4 years ago
#2
(Original post by 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 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 + = 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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#3
(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 + = W
Hence:
Normal = Try again with that value for the normal, and you should get the textbook's answer (I have worked it through).
0
4 years ago
#4
(Original post by Rashina)
Glad I could help 0
#5
(Original post by K-Man_PhysCheM)
Glad I could help have any idea? This is the last one for today.
0
4 years ago
#6
(Original post by Rashina)

have any idea? This is the last one for today.
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.
1
#7
(Original post by 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 can understand the diagram and the answer. Thanks. 0
#8
(Original post by 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
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

0
4 years ago
#9
(Original post by 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
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

0.9g + 0.7g - 1.2g +T - T = 0.4g, NOT 0.8g
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#10
(Original post by K-Man_PhysCheM)

0.9g + 0.7g - 1.2g +T - T = 0.4g, NOT 0.8g
Sorry for that. My bad. Thanks
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#11
(Original post by K-Man_PhysCheM)

0.9g + 0.7g - 1.2g +T - T = 0.4g, NOT 0.8g
no doubts.just ignore the image
0
4 years ago
#12
(Original post by 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.
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. 0
4 years ago
#13
(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!
0
#14
(Original post by K-Man_PhysCheM)
Ah yes, my bad. Thank you for checking!

the c) part
0
#15
(Original post by K-Man_PhysCheM)

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
0
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