Question

Figure 12-50 shows a $\mathbf{70}\mathbf{}\mathit{k}\mathit{g}$climber hanging by only the crimp holdof one hand on the edge of a shallow horizontal ledge in a rock wall. (The fingers are pressed down to gain purchase.) Her feet touch the rock wall at distance$\mathit{H}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.0}\mathit{m}$directly below her crimped fingers but do not provide any support. Her center of mass is distance $\mathit{a}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.20}\mathit{m}$from the wall. Assume that the force from the ledge supporting her fingers is equally shared by the four fingers. What are the values of the(a) horizontal component F_hand (b) vertical component F_vof the force on eachfingertip?

Short answer

a) The horizontal force component of force is$17N$ .

b) The vertical force component of force is $170N$.

Step-by-step solution

Listing the given quantities

The mass of climber is,$m=70kg$.

The climber touches the rock wall at a distance,$H=2m$.

The center of mass is at,$a=0.20m$ .

Understanding the concept of component resolve

We can write the equation of the horizontal component of forces. We can then write the torque equation at point O. using these two equations, we can get the horizontal component of force. To find the vertical force component, use the vertical force equilibrium condition.

Free Body Diagram

Free Body Diagram:

(a) Calculations of the horizontal force component

Using equilibrium condition, the net force in x direction can be written as,

$\sum F_x=0$

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

$\sum F_y=0$

(1)

Using equilibrium condition, the net force in y- direction can be written as,

$\sum F_y=0$

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

$4F_v-mg=0$

(2)

Torque at point acan be calculated as,

$\begin{array}{c}mga-\left(4F_h\right)H=0\\ F_h=\frac{mga}{4H}\end{array}$

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

$\begin{array}{c}{F}_h=\frac{70\text{\hspace{0.17em}kg}\times 9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\times 0.2\text{\hspace{0.17em}m}}{4\times 2.0\text{\hspace{0.17em}m}}\\ =17.16\cdot (\frac{1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}\times 1\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}m}}\times \frac{1N}{1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}})\\ =17N\end{array}$

Thus, the horizontal force component of force is$17N$ .

(b) Calculations of the vertical force component 

From equation 2, we can write,

$F_v=mg/4$

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

$\begin{array}{c}{F}_v=\frac{70\text{\hspace{0.17em}kg}\times 9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}}{4}\\ =171.675\cdot (1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}\times \frac{1N}{1\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}})\\ ~170N\end{array}$

Thus, the vertical force component of force is$170N$$170N$.