91Ó°ÊÓ

First, let's list the forces acting on the ladder: 1. The weight of the ladder (\(W_l = 120 N\)) acts vertically downward at its center of mass, which is at the midpoint of the ladder (\(2.5 m\) from each end). 2. The weight of the painter (\(W_p = 680 N\)) acts vertically downward at the point where the painter stands (\(3.0 m\) above the ground). 3. The force of friction (\(F_f\)) at the bottom of the ladder, which acts horizontally to the right. 4. The normal force (\(N\)) exerted by the wall on the ladder, acting horizontally to the left at the upper end of the ladder (\(4.7m\) above the ground). Since the ladder makes contact with the ground and the wall, we take the distances from these points. Let \(d_w\) be the horizontal distance from the bottom of the ladder to the wall, and \(d_g\) be the vertical distance from the ground to the top of the ladder. Now, we can use the Pythagorean theorem to find \(d_w\) and \(d_g\). The length of the ladder is \(5.0m\), and it leans against the wall at a point \(4.7m\) above the ground, meaning: \(d_w^2 + 4.7^2 = 5.0^2\)

Solve the above equation for \(d_w\): \(d_w^2 = 5.0^2 - 4.7^2\) \(d_w^2 = 25 - 22.09\) \(d_w^2 = 2.91\) \(d_w = \sqrt{2.91} = 1.706\) Therefore, the horizontal distance from the bottom of the ladder to the wall is \(d_w = 1.706m\). We can now calculate the vertical distance, \(d_g\), from the ground to the top of the ladder. Indeed, we already have it: \(d_g = 4.7m\).

To find the force of friction, we will use the conditions for static equilibrium: 1. The net torque acting on the ladder is zero: \(\Sigma \tau = 0\) 2. The net force acting on the ladder in the horizontal direction is zero: \(\Sigma F_x = 0\) 3. The net force acting on the ladder in the vertical direction is zero: \(\Sigma F_y = 0\) By applying these conditions, we will be able to find the force of friction.

Let's set up an equation for the net torque around the bottom of the ladder (i.e., where it contacts the driveway). Note that torque is calculated by multiplying the force acting on the ladder and the perpendicular distance from the pivot point. Taking counterclockwise torques as positive, we have: \(\Sigma \tau = N(4.7) - W_l(2.5) - W_p(3.0) = 0\)

Now, let's set up equations for the net forces in the horizontal (\(F_x\)) and vertical (\(F_y\)) directions: \(\Sigma F_x = F_f - N = 0\) \(\Sigma F_y = W_l + W_p - F_n = 0\)

We can rearrange the equations to solve for the unknown forces: \(F_f = N\) (from the equation for horizontal forces) \(F_n = W_l + W_p = 120 N + 680 N = 800 N\) (from the equation for vertical forces) Now, we can substitute these expressions into the torque equation and solve for the force of friction: \(0 = (F_f)(4.7) - 120(2.5) - 680(3.0)\) \(0 = (F_f)(4.7) - 300 - 2040\) \(F_f = \frac{300 + 2040}{4.7} = \frac{2340}{4.7}\) \(F_f = 497.87 N\) Therefore, the force of friction that the driveway exerts on the bottom of the ladder is approximately \(497.87 N\).