91Ó°ÊÓ

Let's define the following variables: - \(W_1 = 60.0\,\mathrm{lb}\): weight of the boy - \(W_2 = 30.0\,\mathrm{lb}\): weight of the plank - \(L_b = 8.0\,\mathrm{ft}\): total length of the plank - \(L_{s1} = 2.0\,\mathrm{ft}\): distance between the left support and the left end of the plank - \(L_{s2} = 2.0\,\mathrm{ft}\): distance between the right support and the right end of the plank - \(F_1\): force exerted by the left support - \(F_2\): force exerted by the right support (a) If the boy is 3.00 ft from the left end, what force is exerted by each support?

In order for the plank to be in equilibrium, the sum of the vertical forces acting on it should be equal to zero: \(\sum F_y = F_1 + F_2 - W_1 - W_2 = 0\) Now, let's calculate the torque (moment) about the left support, where the distance between the support and the forces \(W_1\), \(W_2\), and the right support \(F_2\) are given:

In order for the plank to be in equilibrium, the sum of the torques about any point should be equal to zero. So let's calculate the torque about the left support (counterclockwise torque is positive and clockwise torque is negative): \(\sum \tau_{left} = W_1 \times (L_{s1}+1) - W_2 \times \frac{L_b}{2} + F_2 \times (L_b - L_{s2}) = 0\) Now, we have two equations and two unknowns \(F_1\) and \(F_2\): 1. \(F_1 + F_2 - W_1 - W_2 = 0\) 2. \(60.0 \times 3 - 30.0 \times 4 + F_2 \times 4 = 0\)

We can now solve these equations for \(F_1\) and \(F_2\). Let's first calculate \(F_2\): \(F_2 = \frac{60.0 \times 3 - 30.0 \times 4}{4} = \frac{180 - 120}{4} = 15.0\,\mathrm{lb}\) Now, substitute \(F_2\) into equation 1 to find \(F_1\): \(F_1 = W_1 + W_2 - F_2 = 60.0 + 30.0 - 15.0 = 75.0\,\mathrm{lb}\) So, the force exerted by the left support is \(75.0\,\mathrm{lb}\) and the force exerted by the right support is \(15.0\,\mathrm{lb}\). (b) The boy moves toward the right end. How far can he go before the plank will tip?

In this case, we need to find the position of the boy before the support on the left stops exerting force and the plank tips. Let's denote the distance of the boy from the left end as \(x\). When the plank is on the verge of tipping, \(F_1 = 0\), and the sum of the torques about the left support should still be equal to zero: \(\sum \tau_{left} = W_1 \times x - W_2 \times \frac{L_b}{2} + F_2 \times (L_b - L_{s2}) = 0\) Now, we can solve for \(x\): \(x = \frac{W_2 \times \frac{L_b}{2} - F_2 \times (L_b - L_{s2})}{W_1} \) Substituting the known values: \(x = \frac{30.0 \times \frac{8}{2} - 15.0 \times (8 - 2)}{60.0} \) \(x = \frac{120 - 90}{60} = \frac{30}{60} = 0.5\,\mathrm{ft}\) So, the boy can move 0.5 ft to the right from his original position (3.00 ft from the left end) before the plank will tip. This means he can walk up to 3.5 ft from the left end.