Equilibrium of particles

• Dec 23rd 2009, 08:23 AM
BabyMilo
Equilibrium of particles
Quote:

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.

Quote:

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.
Quote:

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
$\rightarrow$ 5=0.2(mg)
$\rightarrow$ m=2.55kg

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

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

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

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

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

is this right? thanks!
• Dec 24th 2009, 10:55 AM
Hello BabyMilo
Quote:

Originally Posted by BabyMilo
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
$\rightarrow$ 5=0.2(mg)
$\rightarrow$ m=2.55kg

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

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

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

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

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

is this right? thanks!

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

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

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

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

$\Rightarrow X = 39.8^o$