Aim: More Law of Acceleration Solve for the normal force on the following diagram: 10 kg 2 N N mg ΣF = 0 2 N + N – mg = 0 N = mg – 2 N N = (10 kg)(10 m/s 2 ) – 2 N N = 98 N Answer Key HW 5

a. On the diagrams to the right, draw and label the forces acting on the hook and the forces acting on the load as they accelerate upward b. Determine the tension T 1 in the lower cable and the tension T 2 in the upper cable as the hook and load are accelerated upward at 2 m/s 2. Use g = 10 m/s². 1. A crane is used to hoist a load of mass m 1 = 500 kilograms. The load is suspended by a cable from a hook of mass m 2 = 50 kilograms, as shown in the diagram above. The load is lifted upward at a constant acceleration of 2 m/s 2. T2T2 T1T1 m2gm2g T1T1 m1gm1g Calculator **5 min**

ΣF = ma T 1 – m 1 g = m 1 a T 1 = m 1 a + m 1 g T 1 = (500 kg)(10 m/s 2 ) + (500 kg)(2 m/s 2 ) T 1 = 5,000 N + 1,000 N T 1 = 6,000 N ΣF = ma T 2 – T 1 – m 2 g = m 2 a T 2 = m 2 a + T 1 + m 2 g T 2 = (50 kg)(2 m/s 2 ) + (6,000 N) + (50 kg)(10 m/s 2 ) T 2 = 100 N + 6,000 N N T 2 = 6,600 N

2. A student of mass m stands on a platform scale in an elevator in a tall building. The positive direction for all vector quantities is upward. a. Draw a free-body diagram showing and labeling all the forces acting on the student, who is represented by the dot below. N mg Calculator **11 min**

b. Derive an expression for the reading on the scale in terms of the acceleration a, the mass m on the student, and fundamental constants. ΣF = ma N – mg = ma N = ma + mg N = m (a + g)

An inspector provides the student with the following graph showing the acceleration a of the elevator as a function of time t.

c. i. During what time interval(s) is the force exerted by the platform scale on the student a maximum value? The normal force is maximum when the acceleration is maximum The force exerted by the platform scale on the student is maximum from 12 s to 16 s.

ii. Calculate the magnitude of that maximum force for a 45 kg student. N = m(a + g) N = 45 kg(1.2 m/s m/s 2 ) N = 45 kg(11 m/s 2 ) N = 500 N d. During what time interval(s) is the speed of the elevator constant? Speed is constant when the acceleration is zero From 7 s to 10 s and from 17 s to 20 s

3. A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling, as shown above. One end of the rope is held by Student A of mass 70 kg, who is at rest on the floor. The opposite end of the rope is held by Student B of mass 60 kg, who is suspended at rest above the floor. Calculator **15 min**

a. On the dots below that represent the students, draw and label free-body diagrams showing the forces on Student A and on Student B.

b. Calculate the magnitude of the force exerted by the floor on Student A. ΣF A = 0 ΣF B = 0 T + N – m A g = 0T – m B g = 0 m B g = T T + N – m A g = 0 m B g = T N + m B g – m A g = 0 N = m A g – m B g N = g(m A – m B ) N = 9.8 m/s 2 (70 kg – 60 kg) N = 98 N

Student B now climbs up the rope at a constant acceleration of 0.25 m/s 2 with respect to the floor. c. Calculate the tension in the rope while Student B is accelerating. ΣF B = m B a T – m B g = m B a T = m B a + m B g T = m B (a + g) T = 60 kg (0.25 m/s m/s 2 ) T = 603 N

d. As student B is accelerating, is Student A pulled upward off the floor? Justify your answer. No. To lift the student off the floor, the tension must be greater than the student’s weight. e. With what minimum acceleration must Student B climb up the rope to lift Student A upward off the floor?