Homework #2 Solutions
Physics 221 Summer 2012
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
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