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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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    (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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    (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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    (Original post by Rashina)
    I got the answer. Thanks
    Glad I could help
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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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    (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: 61
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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    (Original post by K-Man_PhysCheM)
    Here is a terrible diagram:
    Name:  forcediagram.png
Views: 61
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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    (Original post by K-Man_PhysCheM)
    Here is a terrible diagram:
    Name:  forcediagram.png
Views: 61
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}


    Name:  Mechanics 4.jpg
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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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    (Original post by Rashina)
    Name:  Mechanics 4.jpg
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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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    (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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    (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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    (Original post by K-Man_PhysCheM)
    Here is a terrible diagram:
    Name:  forcediagram.png
Views: 61
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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    (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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    (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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    (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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