Figure 7 shows two of the forces acting on a uniform ladder resting against a vertical wall - AQA - A-Level Physics - Question 5 - 2017 - Paper 1

An arrow should be drawn vertically upward from the point of contact between the ladder and the ground to indicate the resultant force from the ground acting on the ladder. Label this arrow 'G'.

Using the sum of clockwise moments equals the sum of anticlockwise moments, we can calculate the resultant force. The weight acts at the center of the ladder, which is at a distance of 4.0 m from the ground contact point. Thus, the moment due to the weight is:

390 imes 4.0 imes rac{1}{2}= 110 \text{ Nm}

Now considering the angle of the ladder, we have:

$F_w \cdot 8.0 \cdot \sin(60) = 390 \cdot 4.0 \cdot \cos(60)$

Solving this gives:

$F_w = \frac{390 \cdot 4.0}{8.0 \cdot \sin(60)}$

This results in:

$F_w \approx 113 \text{ N}$

When someone climbs the ladder, the vertical reaction force from the ground increases due to the additional weight of the person. The direction of the resultant force from the ground also changes as the distribution of forces is altered. The frictional force at the wall may also be affected; as the upward force increases, the component of weight from the wall would cause the ladder to push harder against the wall, increasing the friction element.