91影视

The condition for translational equilibrium is: $\mathbf{鈭�}{\mathit{F}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}$ And that for rotational equilibrium is: $\mathbf{鈭�}{\mathit{蟿}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}$. The total of the forces will be zero.

(a)

The ladder, placed against a frictionless wall, is at equilibrium. The wall exerts a normal force of $N_1\text{Newton}$, and the normal force exerted by the ground is $N_2\text{Newton}$.

Let the whole given setup be illustrated as a free body diagram for forces on the ladder, as shown in the figure as:

Here, the weight of the ladder and that of man are, respectively $W_l\text{and}{W}_m$.

Now, the frictional force applied by the ground on the ladder will be:

$\begin{array}{rcl}F& =& {渭}_s{N}_2\\ F& =& 0.40N_2\\ & & \end{array}$ (1)

Since the ladder is at equilibrium, the weights and the forces will be related as:

$\begin{array}{rcl}{N}_2& =& W_l+W_m\\ & =& 160+740\\ & =& 900\text{N}\\ & & \end{array}$

Put this value in equation (1), we get:

$\begin{array}{rcl}F& =& 0.40N_2\\ & =& 0.40脳900\\ & =& 360\text{N}\\ & & \end{array}$

Thus, the friction force applied by the ground is$360\text{N}$ .

(b)

Applying the condition for rotational equilibrium, the torque on the ladder can be calculated as:

$\begin{array}{rcl}鈭�蟿& =& 0\\ F_{\text{wall}}路\left(4\right)& =& \left(1.5\right)W_l+\left(1\right){渭}_s{W}_m\\ 4F_{\text{wall}}& =& 1.5脳160+1脳0.6脳740\\ F_{\text{wall}}& =& 171\text{N}\end{array}$

Thus, the friction force when the man had climbed 1 meter was 171 Newtons.

(c)

Assuming that the man has climbed the distance xmeters, the torque just before the ladder started to slip can be calculated as:

$\begin{array}{rcl}鈭�蟿& =& 0\\ N_2\left(4\right)& =& \left(1.5\right)W_l+\left(x\right){渭}_s{W}_m\\ 4脳360& =& 1.5脳160+\left(x\right)脳0.6脳740\\ x& =& 2.7\text{m}\\ & & \end{array}$

Thus, the ladder started to slip when the man climbed the ladder to the length of 2.7 meters.