HOMEWORK #2

Due Friday June 22, 2012

1

A 70.0-kg person stands on a scale placed on the ﬂoor of an elevator. Find: - the weight of the person (magnitude and direction), - the normal force by the scale on the person (magnitude and direction), - and what the scale reads (in kilograms) in the following cases: (a) The elevator moves up with a constant speed of 2.0 m/s2 . (b) The elevator has a constant upward acceleration of 2.0 m/s2 . (c) The elevator has a constant downward acceleration of 2.0 m/s2 . (d) The cable snaps and the elevator falls freely (ignore friction and the bloody end!). SOLUTION: The weight of the person depends on Mass and g, therefore: W = M g = (70.0 kg) 9.80 m/s2 = 686 N downward in all cases (a)-(d). Using the free-body-diagram for the person is shown to the right, We solve for N (which equals to what the scale reads) using Newton’s second law of motion: F = Ma N − Mg = Ma ⇒ N = M a + M g = M (a + g) (1) Mg N

(a) Constant speed → a = 0, then equation 7 reads: N = M (0 + g) = M g = 686 N upward (b) a = m/s2 upward. Equation 7 now reads: N = M 2.0 m/s2 + 9.80 m/s2 = 826 N upward (c) a = m/s2 downward. Equation 7 now reads: N = M −2.0 m/s2 + 9.80 m/s2 = 546 N upward (d) Free fall → a = −g: N = M (−g + g) = 0

1

Physics 221 Summer 2012

HOMEWORK #2

Due Friday June 22, 2012

2

Three boxes labeled A, B and C with masses m, m and M (M > m), respectively, are attached with ideal massless strings as shown in the ﬁgure below. The string goes through an ideal pulley. Neglect friction and air resistance. Rank from largest to smallest and explain: (a) The acceleration of each box and the acceleration of gravity. (b) The net force on each box. (c) The tension in each string and the weight of box C. SOLUTION: The free-body diagrams are drawn on the ﬁgure to the right. (a) The acceleration of all boxes is the same. Their acceleration is less than g. aA = aB = aC < g (b) The net force on an objects A and B is ma and on C is M a, so: Fnet,A = Fnet,B < Fnet,C (c) For Object A: T1 = ma For Object B: T2 − T1 = ma ⇒ T2 = 2ma For Object C: M g − T 2 = M a ⇒ M g = (M + 2m) a Therefore: M g > T2 = 2T1 Mg C M m T1 A m B T2 T2

2

Physics 221 Summer 2012

HOMEWORK #2

Due Friday June 22, 2012

3

Consider the two blocks at rest on a horizontal surface shown below. Discuss the following statements (True/False and why): (a) The net force on either block is zero. (b) There is a normal force on 1 with magnitude m1 g and direction up. (c) The weight of block 1 acts on block 2. (d) The force on block 1 by block 2 is twice as large as the force on block 2 by block 1. (e) The magnitude of the normal force by the ﬂoor on block 2 is equal to m2 g. SOLUTION: (a) True; Since neither box accelerate, then the net force must be zero. (b) True; The normal force is equal and opposite to all other forces acting block 1 downward, which in this case only m1 g downward. (c) False; The weight of an object is the gravitational force of the planet acting on that object. (d) False; According to newton’s third law, action and reaction forces are equal in magnitude. (e) False; The ﬂoor is supporting m1 + m2 . Therefore the Normal force of the ﬂoor on m2 is Nﬂoor on block 2 = (m1 + m2 ) g. 1 m2 = 2m1 2

3

Physics 221 Summer 2012

HOMEWORK #2

Due Friday June 22, 2012

4

An 80 kg person is parachuting and experiencing a downward acceleration of 2.5 m/s2 . The mass of the parachute is 5.0 kg. (a) What is the upward force on the open parachute from the air? (b) What is the downward force on the parachute from the person? SOLUTION: The free-body diagram on the person+parachute system is shown to the right: (a) On person+parachute: Fext = (M + m) a Fair − (M + m) g = (M + m) a ⇒ Fair = (M + m) (a + g) = (85.0 kg) −2.5 m/s2 + 9.80 m/s2 = 620.5 N upward where m was used to denote parachute’s mass, and M to denote person’s mass. (b) On parachute only (see diagram to the right): Fon parachute...