Question

To get up on the roof, a person (mass 80.0 kg) places a 5.40 m aluminum ladder (mass 13.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes (in N) of the forces on the ladder at the top and bottom?TOP =BOTTON =

Answer 1

The given problem can be exemplified in the following diagram:

Where:

$\begin{gathered} Ft=\text{ force at the top} \\ W_p=\text{ weight of the person} \\ W_l=\text{ weight of the ladder} \\ F_{by},F_{bx}=\text{ y and x components of the force at the bottom} \end{gathered}$

To determine the value of the forces at the top and bottom we will first use the sum of moments with respect to point B. The moments in the counterclockwise direction will be considered to be positive.

The moments are:

$\Sigma M_B=W_p(3\cos\theta)+W_l(2\cos\theta)-F_Th$

Since the ladder is not turning this means that the sum of moments is zero, therefore, we have:

$W_p(3\cos\theta)+W_l(2\cos\theta)-F_Th=0$

Now, since the weight is the product of the mass and the acceleration of gravity we have:

$m_pg(3\cos\theta)+m_lg(2\cos\theta)-F_Th=0$

Where:

$\begin{gathered} m_p=\text{ mass of the person} \\ m_l=\text{ mass of the ladder} \end{gathered}$

We can use the right triangle formed by the ladder and the wall to determine the cosine of the angle theta, like this:

Since the cosine is defined as:

$\cos\theta=\frac{adjacent}{hypotenuse}$

Substituting we get:

$\cos\theta=\frac{2}{5.4}$

Now, we substitute the value of the cosine:

$m_pg(3(\frac{2}{5.4}))+m_lg(2(\frac{2}{5.4}))-F_Th=0$

To determine the value of "h" we apply the Pythagorean theorem to the right triangle:

$2^2+h^2=5.4^2$

Now, we solve the squares:

$4+h^2=29.16$

Now, we subtract 4 from both sides:

$\begin{gathered} h^2=29.16-4 \\ h^2=25.16 \end{gathered}$

Now, we take the square root to both sides:

$h=\sqrt{25.16}$$h=5.02$

Now, we substitute the values:

$(80kg)(9.8\frac{m}{s^2})(3m(\frac{2}{5.4}))+(13kg)(9.8\frac{m}{s^2})(2m(\frac{2}{5.4}))-F_T(5.02m)=0$

Now, we solve the operations:

$965.48Nm-F_t(5.02m)=0$

Now, we solve for the force at the top. Adding Ft(5.302m) to both sides:

$965.48Nm=F_t(5.02m)$

Now, we divide both sides by 5.02m:

$\frac{965.48Nm}{5.02m}=F_t$

Solving the operations:

$192.33N=F_t$

Now, to determine the forces at the bottom we will use the sum of forces. For the x-component, we add the horizontal forces, like this:

$\Sigma F_h=0$

Adding the forces we get:

$F_t+F_{bx}=0$

Now, we subtract the force at the bottom from both sides:

$F_{bx}=-F_t$

Now, we substitute:

$F_{bx}=-192.33N$

Now, we add the vertical forces:

$F_{by}-m_pg-m_lg=0$

Now, we solve for the y-component:

$F_{by}=m_pg+m_lg$

Now, we plug in the values:

$F_{by}=(80kg)(9.8\frac{m}{s^2})+(13kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_{by}=911.4N$

Now, the magnitude of the bottom force is given by:

$F_b=\sqrt{F_{bx}^2+F_{by}^2}$

Substituting the values we get:

$F_b=\sqrt{(-192.33N)^2+(911.4N)^2}$

Solving the operations:

$F_b=931.47N$

Therefore, the bottom force is 931.47N