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Thread: Equilibrium of particles

  1. #1
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    Equilibrium of particles

    A block of mass m kg is at rest on a horizontal plane. The coefficient of friction between the block and the plane is 0.2.

    i) When a horizontal force of magnitude 5N acts on the block, the block is on the point of slipping. Find tht value of m.
    ii) When a force of magnitude P N acts downwards on the block at angle X to the horizontal, as shown in the diagram, the frictional force on the block has magnitude 6N and the block is again on the point of slipping. Find P and X.
    I just need to check if im right because the mark scheme is not available anymore on the exam board website. THanks.

    i) F=uR
    $\displaystyle \rightarrow$ 5=0.2(mg)
    $\displaystyle \rightarrow$ m=2.55kg

    ii)
    $\displaystyle fx= PcosX-6=0$
    $\displaystyle fy= R-mg-PsinX=0$

    so $\displaystyle \rightarrow$ P=$\displaystyle \frac{6}{cosX}$

    R=$\displaystyle \frac{6}{0.2}=30N$

    $\displaystyle 30-(\frac{6}{cosX})(sinX)-2.55g=0$
    $\displaystyle \rightarrow$$\displaystyle -6tanX=-5$
    $\displaystyle \rightarrow$$\displaystyle tanX=\frac{5}{6}$
    $\displaystyle X=39.8 (3s.f.)$

    $\displaystyle P=\frac{6}{cos39.8}=7.81N (3s.f.)$

    is this right? thanks!
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  2. #2
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    Hello BabyMilo
    Quote Originally Posted by BabyMilo View Post
    I just need to check if im right because the mark scheme is not available anymore on the exam board website. THanks.

    i) F=uR
    $\displaystyle \rightarrow$ 5=0.2(mg)
    $\displaystyle \rightarrow$ m=2.55kg

    ii)
    $\displaystyle fx= PcosX-6=0$
    $\displaystyle fy= R-mg-PsinX=0$

    so $\displaystyle \rightarrow$ P=$\displaystyle \frac{6}{cosX}$

    R=$\displaystyle \frac{6}{0.2}=30N$

    $\displaystyle 30-(\frac{6}{cosX})(sinX)-2.55g=0$
    $\displaystyle \rightarrow$$\displaystyle -6tanX=-5$
    $\displaystyle \rightarrow$$\displaystyle tanX=\frac{5}{6}$
    $\displaystyle X=39.8 (3s.f.)$

    $\displaystyle P=\frac{6}{cos39.8}=7.81N (3s.f.)$

    is this right? thanks!
    Good job! I agree with all your answers.

    For number 2, I started the same way as you:

    Resolve horizontally:
    $\displaystyle fx= P\cos X-6=0$
    $\displaystyle \Rightarrow P \cos X = 6$ (1)
    Resolve vertically:
    $\displaystyle fy= R-mg-P\sin X=0$ (2)
    Friction is limiting, so:
    $\displaystyle R=\frac{6}{0.2}=30$
    I then said, from part 1:
    $\displaystyle mg = \frac{5}{0.2}= 25$
    Substitute into (2):
    $\displaystyle 30-25-P\sin X = 0$

    $\displaystyle \Rightarrow P\sin X = 5$ (3)
    Square and add (1) and (3):
    $\displaystyle P^2 = 36+25$

    $\displaystyle \Rightarrow P = \sqrt{61}= 7.81$
    Divide (3) by (1):
    $\displaystyle \tan X = \frac{5}{6}$

    $\displaystyle \Rightarrow X = 39.8^o$
    Grandad
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