91影视

• Mass of the person, m = 49kg.
• Coefficient of friction between shoes and rock, ${渭}_{s1}=1.2$.
• Coefficient of friction between back and rock, ${渭}_{s2}=0.8$.

The problem deals with Newton鈥檚 second law of motion, which states that an object's acceleration depends on the net force acting upon the object and the mass of the object. First, draw the free body diagram of the person. Then, solve the given problem using Newton鈥檚 second and third laws.

The free-body diagram for the person (shown as an L-shaped block) is shown below.

The force she exerts on the rock slabs is not directly shown (since the diagram should only show forces exerted on her). Still, Newton's third law relates it) to the normal forces ${\stackrel{鈬赌}{F}}_{N1}$and ${\stackrel{鈬赌}{F}}_{N2}$exerted horizontally by the slabs onto her shoes and back, respectively.

We apply Newton's second law to the x and y axes (with +x rightward and +y upward, and there is no acceleration in either direction).

$\begin{array}{rcl}{F}_{N1}-F_{N2}& =& 0\\ f_1+f_2-mg& =& 0\end{array}$

The first equation tells us that the normal forces are equal $F_{N1}=F_{N2}=F_N$. Consequently, we know that, the friction force is given by,

$$\begin{gathered} f_1={渭}_{s1}{F}_N \\ {f}_2={渭}_{s2}{F}_N \end{gathered}$$

We conclude that,

$f_1=\left(\frac{{渭}_{s1}}{{渭}_{s2}}\right)f_2$

Therefore,

$$\begin{gathered} f_1+f_2-mg=0 \\ \left(\frac{{渭}_{s1}}{{渭}_{s2}}+1\right)f_2=mg \end{gathered}$$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}\left(\frac{1.2}{0.8}+1\right)f_2& =& 49kg脳9.8m/s^2\\ f_2& =& 192N\end{array}$

From this, we can find the force normal force as:

$\begin{array}{rcl}{F}_N& =& f_2/{渭}_{s2}\\ & =& \frac{192N}{0.80}\\ & =& 240N\end{array}$

This is equal to the magnitude of the push exerted by the rock climber.

Thus, the magnitude of the force is 240 N.

From the above calculation, we find force as:

$\begin{array}{rcl}{f}_1& =& {渭}_{s1}{F}_N\\ & =& 1.2脳240N\\ & =& 288N\end{array}$

The fraction can be calculated as:

$\begin{array}{rcl}\frac{f_1}{W}& =& \frac{288}{\left(49\right)\left(9.8\right)}\\ & =& 0.60\end{array}$

or 60% of her weight.

Thus, 60% or .60 is the fraction of her weight that is supported by frictional force on her shoes.