3.3) I'm not sure whether it is correct or not. I'm am confused because of the Applied Force being at an angle. Am I only working out Fx and would...

Thursday, October 4, 2007

3.3) I'm not sure whether it is correct or not. I'm am confused because of the Applied Force being at an angle. Am I only working out Fx and would...

Hello!

There are some mistakes in your solution. First, the transition from $\displaystyle{F}_{{f{}}}$ to $\displaystyle\mu\cdot{m}{g{}}$ is incorrect. We know that $\displaystyle{F}_{{f{=}}}\mu\cdot{N},$ where $\displaystyle{N}$ is the reaction force. But in this case $\displaystyle{N}\ne{m}{g{}}$ because $\displaystyle{F}$ acts partly upwards. We have to consider the both projections, vertical and horizontal.

For the vertical axis we obtain $\displaystyle{N}-{m}{g{+}}{F}\cdot{\sin{{\left(\alpha\right)}}}={0}$

and from the horizontal $\displaystyle{F}\cdot{\cos{{\left(\alpha\right)}}}-{F}_{{{f}}}={F}\cdot{\cos{{\left(\alpha\right)}}}-\mu\cdot{N}={0}.$

Express $\displaystyle{N}$ from the first equation and substitute it into the second:

$\displaystyle{N}={m}{g{-}}{F}{\sin{{\left(\alpha\right)}}},$

$\displaystyle{F}{\cos{{\left(\alpha\right)}}}=\mu{\left({m}{g{-}}{F}{\sin{{\left(\alpha\right)}}}\right)}=\mu{m}{g{-}}{F}\mu{\sin{{\left(\alpha\right)}}}.$

So $\displaystyle{F}{\left({\cos{{\left(\alpha\right)}}}+\mu{\sin{{\left(\alpha\right)}}}\right)}=\mu{m}{g{}}$ and the final formula is

$\displaystyle{F}=\frac{{\mu{m}{g}}}{{{\cos{{\left(\alpha\right)}}}+\mu{\sin{{\left(\alpha\right)}}}}}.$

Now recall that $\displaystyle\alpha$ =20°, $\displaystyle\mu={0.4},$ $\displaystyle{m}$ =50 kg and $\displaystyle{g{}}$ =9.8 $\displaystyle\frac{{m}}{{{s}}^{{2}}},$ and compute:

$\displaystyle{F}\approx\frac{{{0.4}\cdot{50}\cdot{9.8}}}{{{0.94}+{0.4}\cdot{0.34}}}\approx\frac{{196}}{{1.08}}\approx$ 181 (N). This is the answer for 3.3. Now you can easily solve 3.4 and 3.5.

Note also that 0.4*50*9.8 = 196 and not 16 as you wrote (maybe you simply omitted 9).

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Thursday, October 4, 2007

3.3) I'm not sure whether it is correct or not. I'm am confused because of the Applied Force being at an angle. Am I only working out Fx and would...

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What was the device called which Faber had given Montag in order to communicate with him?