Question

Interactive Exercises 5.14: Applying Newton's Laws (The Double Incline) In a project for a mechanical engineering class, some students make a double incline, in which the angle of the two inclined, smooth surfaces can be separately adjusted by means of a hinge at the top-see Fig. 5.14.1. The angle of the incline on the left side is denoted by the angle φ shown in the figure. The angle of the right incline is denoted by the angle θ. Two blocks, with smooth bases to minimize the effects of friction, are placed on the inclines, one on each side. A massless string connects the blocks and passes over an idealized pulley. The mass of block A is mA-1.00 kg. You can see a representation of this setup in the simulation (linked below) Interactive Figure 5.14.1: A double incline consists of two inclined, smooth surfaces, the angles of which can be adjusted separately. One block is placed on each side of the double incline, and the blocks are connected by a massless string that passes over an idealized pulley Question 1 In their first experiment with the apparatus, the students use the double incline to measure the unknown mass of block B. They do this by fixing the value of the right incline at θ-20.00 and then adjusting the value of φ for the left incline such that both blocks remain at rest. They find the left incline must be tilted such that φ = 43.0°. (You can do this experiment for yourself with the simulation!) What is the mass of block B, mB? kg the tolerance is +/-2% By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor Attempts: 0 of 5 used SAVE FOR LATER SUBMIT ANSWER

Answer 1

Here, we use newton,s laws to solve both the parts. For part A, we use the equilibrium condition. For part B we use Fnet = ma

T = 1 g sin phi = m_b g sin theta rightarrow m_b = 1 sin 43 degree /sin 20 degree = 1.994 kg for 1 kg t = 1 a for m_b m_b g sin theta - t = m_b a rightarrow a = m_b g sin theta /1 + m_b = (1.994) (9.8) sin 20/1 + 1.994 = 2.232 m/s^2