Worksheet 4a: Forces

Worksheet 5: Forces

Problem: Weight, Normal force, and a Box
A box of mass 10.0-kg rests on a smooth table as show in the figure.

a) Determine the weight of the box and the normal force acting on it.
STEP 1: Draw the free-body diagram

                    STEP 2: The system is in equilibrium, therefore SF = 0
                            F_N - mg = 0
                            F_N = mg = (10.0 kg)(9.80 m/s²) = 98.0 N

b) You push down on the box with a force of 40.0 N. Now determine the normal force acting on the block.
STEP 1: Draw the free-body diagram

                    STEP 2: The system is in equilibrium, therefore SF = 0
                             F_N - mg - 40.0 N= 0
                             F_N = 98.0 N + 40.0 N = 138 N
c) If you now try to pick up the box with a force of 40.0 N, what would the normal force be now?
                   STEP 1: Draw the free-body diagram

                    STEP 2: The system is in equilibrium, therefore SF = 0
                             F_N - mg + 40.0 N= 0
                             F_N = mg - 40.0 N= 98.0 N - 40.0 N = 48.0 N

Problem 1: Solve the following force problems
[A] What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s²?
F = ma = (60.0 kg)(1.15 m/s²) = 69.0 N

[B] A net force of 225 N accelerates a bike and rider at 2.20 m/s². What is the combined mass of the bike and rider?
F = ma , therefore m = F/a
m = (225 N)/(2.20 m/s²) = 102 kg

Problem 2: Forces of 10.0 N north, 20.0 N east, and 15.0 N south are simultaneously applied to a 4.00-kg mass. Find the acceleration of the mass (both magnitude and direction).

Solution:
                   The vector equation for the forces acting on the object is:
                           F_net = ( 10.0j+20.0i - 15.0j)N
                                     = (20.0i - 5.0j)N
                            |F| = [(20.0)² + (-5.0)²]^1/2 = 20.6 N
                           q = tan^-1[(-5.0)/(20.0)] = -14.0^o

Problem 3: Calculate the following:
[A] A person weighs 125 lb. Determine his weight in Newton’s and her mass in kilograms.(1 lb. = 4.448 N)
F = 125 lb (4.448 N/lb) = 556 N

[B] What is the weight of a 66-kg astronaut on earth?
F_g = mg = (66 kg)(9.80 m/s²) = 647 N

[C] What is the weight of the same 66-kg astronaut on the moon? (g = 1.7 m/s²)
F_g = mg = (66 kg)(1.7 m/s²) = 112 N

[D] What is the weight of the same astronaut in outer space traveling with constant velocity?
Since the astronaut is traveling with constant velocity, he is weightless: F_g = 0

Problem 4: A box weighing 70 N rests on a table. A rope attached to that box runs up over a pulley and a second box is attached to the other end. Determine the Normal force on the box on that table if the hanging box weighs (a) 30 N, (b) 60 N, and (c) 90 N.

                    For the 70 N block: Since SF = 0, then T + F_N - 70 N = 0
                    F_N = 70 N - T
                    For the hanging block: Since SF = 0, then T = F_g
                    Therefore:   F_N = 70 N - F_g
a)   F_g= 30 N
                    F_N = 70 N - F_g= 70 N - 30 N = 40 N
b) F_g= 60 N
                    F_N = 70 N - F_g= 70 N - 60 N = 10 N
c) F_g= 90 N
                    F_N = 70 N - F_g= 70 N - 90 N = -20 N =0 The block has lifted off the table, therefore the normal force is equal to zero.

Problem 5: Two blocks on a horizontal surface are in contact with each other as shown. The surface has friction. A force F is applied to block 1.

(a) Draw a free body diagram for each block: Note: F_N1 and F_N2 are the normal forces exerted by the surface on block 1 and block 2 respectively, F_R is the reaction force caused by the blocks being in contact with each other, f_k1 and f_k2 are the fricional forces associated with block 1 and 2 respectively.

(b) calculate the acceleration of the system (in terms of m₁ and m₂).
Solution:
                   Block 1: Block 1 is in equilibrium in the y-direction but accelerating in the x-direction, therefore:
                           SF_x = m₁a                               SF_y = 0
                            F - F_R - f_k1 = m₁a                    F_N1 = m₁g
                            f_k1= m_kF_N1     therefore         f_k1= m_km₁g
                   Block 2: Block 2 is in equilibrium in the y-direction but accelerating in the x-direction, therefore:
                           SF_x = m₂a                               SF_y = 0
                            F_R - f_k2 = m₂a                        F_N2 = m₂g
                            f_k2= m_kF_N2     therefore         f_k2= m_km₂g
                    This gives usthe following equations:
                            F - F_R - f_k1 = m₁a    (1)
                            f_k1= m_km₁g              (2)
                            F_R - f_k2 = m₂a         (3)
                            f_k2= m_km₂g             (4)
                    Combining (1) and (2) gives us
                            F - F_R - m_km₁g = m₁a    (5)
                    Combining (3) and (4) gives us
                            F_R = m_km₂g + m₂a        (6)
                    Combining (5) and (6) gives us
                            F - (m_km₂g + m₂a ) - m_km₁g = m₁a
                    Now we can finally solve for a:
                           a = [F - m_kg(m₁+m₂)]/(m₁+m₂)

(c) Calculate the net force on each block,
                    Block 1:
                           SF_x = m₁a = m₁{[F - m_kg(m₁+m₂)]/(m₁+m₂)}
                    Block 2:
                           SF_x = m₂a = m₂{[F - m_kg(m₁+m₂)]/(m₁+m₂)}

(d) calculate the contact force that each box exerts on it’s neighbor.
F_R = m_km₂g + m₂[F - m_kg(m₁+m₂)]/(m₁+m₂)

Problem 6: A force of 40.0 N is needed to move a 5.0-kg box across a horizontal concrete floor.
(a) What is the coefficient of static friction between the box and the floor, just before the box moves?

Solution:
                    The system is in equilibrium, therefore:
                           SF_x =0                               SF_y = 0
                            40.0 N = f_s                        F_N = mg = 49.0 N
                            f_s= m_sF_N
                          m_s= f_s/F_N=(40.0 N)/(49.0 N) = 0.816

(b) If the 40.0 N force continues to act on the box, the box accelerates at 0.70 m/s². What is the coefficient of
kinetic friction?
Solution:
                    The system is in not in equilibrium in the x-direction, therefore:
                           SF_x =ma                               SF_y = 0
                            40.0 N - f_k = ma                  F_N = mg = 49.0 N
                            f_k= m_kF_N
                          m_k= f_k/F_N=(40.0 N - 3.50 N)/(49.0 N) = 0.745

Problem 7: A hockey puck is on a frozen pond. It is hit and given an initial speed of 25.0 m/s. If the puck remains on the ice at all times and comes to rest after traveling 115 m , determine the coefficient of kinetic friction between the puck and the ice.
Solution:
                    The system is not in equilibrium because it is accelerating, so first, we have to find the value of the acceleration. Knowns: Dx = 115 m, v_o = 25.0 m/s, v = 0, therefore
                            v² = v_o² - 2aDx
                            a = (v² - v_o²)/2Dx = -2.72 m/s²
                            -f_k =ma
                            m_k= -ma/mg = -a/g = 2.72/9.80 = 0.277

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