Worked Examples: Translational Equilibrium
Subject: Physics | Grade: 11 | Topic: Newton's Laws of Motion
Worked Examples: Translational Equilibrium
Practice problems involving objects moving at constant velocity, objects on inclined planes, and hanging systems in translational equilibrium.
If an object moves at a constant velocity as shown below, find the friction force acting on it.
Since velocity is constant, acceleration is zero and the net force acting on the object is zero.
Fnet = 0
Fapp − Ff = 0
20 − Ff = 0
Answer: Ff = 20 N (left)
If block A moves at a constant speed due west, find the friction force acting on it.
Answer: Ff = 10 N (east)
If an object slides downhill at a constant speed, calculate the friction force acting on it.
Since v = constant, a = 0, therefore Fnet = 0.
Fg∥ − Ff = 0
Ff = mg sinα = 2 × 9.8 × sin30° = 9.8 N
Answer: Ff = 9.8 N (uphill)
If an object slides downhill at a constant speed, calculate the friction force acting on it.
Answer: Ff = 31.31 N (uphill)
If the object is in equilibrium, calculate the magnitude of the tension on the rope.
Since it is in equilibrium: Fnet = 0
Fg − FT = 0
FT = Fg = mg
FT = 5 × 9.8 = 49 N
Answer: FT = 49 N
If the object is in equilibrium, estimate the magnitude of the tension on the rope.
Answer: 98 N
If the blocks are in equilibrium, calculate the magnitudes of the tension on the ropes.
Since the system is in equilibrium, the net force on each mass is zero.
FT1 = m1g = 5 × 9.8 = 49 N
FT2 = m2g + FT1 = 3 × 9.8 + 49 = 78.4 N
Answer: FT1 = 49 N, FT2 = 78.4 N
If the object is in equilibrium, estimate the magnitudes of the tension on the ropes.
Answer: FT1 = 49 N, FT2 = 98 N
If the blocks are in equilibrium, calculate the magnitudes of the tensions on the ropes.
Because the system is in equilibrium, horizontal forces cancel and vertical forces balance the weight.
FT2∥ = FT1∥, so FT1 = FT2 = A
FT1⊥ + FT2⊥ = Fg
A cos45° + A cos45° = mg
2A cos45° = 10 × 9.8
Answer: FT1 = FT2 = 69.3 N
If the system is in equilibrium, find the magnitudes of the tension on the ropes.
Answer: FT1 = 42.44 N, FT2 = 24.5 N
Lift Problems
Suppose there is a scale on the floor of a lift and a person is standing on it. When the lift starts to move, the reading on the scale may change with respect to the motion. The reading on the scale is also called the apparent weight of the person.
1.1 is at rest
1.2 moves downward at constant velocity of 2 m·s−1
1.3 moves upward at constant velocity of 2 m·s−1
1.4 accelerates upward at 2 m·s−2
1.5 accelerates upward at −2 m·s−2
1.6 accelerates downward at 3 m·s−2
1.7 accelerates downward at −3 m·s−2
1.8 falls freely when the cable breaks
A man with a mass of 80 kg stands on a bathroom scale on the floor of a lift. Determine the reading on the scale in each case.
A lift is in equilibrium when it is at rest or moving up or down with constant velocity. In those cases, acceleration is zero and the apparent weight is equal to the actual weight.
Fnet = FN − Fg = ma
FN = apparent weight, Fg = actual weight
1.1 At rest:
a = 0, so FN = Fg = mg = 80 × 9.8 = 784 N
1.2 Moving downward at constant velocity:
a = 0, so FN = 784 N
1.3 Moving upward at constant velocity:
a = 0, so FN = 784 N
1.4 Accelerating upward at 2 m·s−2:
FN − Fg = ma
FN = ma + mg = 80 × 2 + 80 × 9.8 = 944 N
1.5 Accelerating upward at −2 m·s−2:
FN = ma + mg = 80 × (−2) + 80 × 9.8 = 624 N
1.6 Accelerating downward at 3 m·s−2:
Fg − FN = ma
FN = mg − ma = 80 × 9.8 − 80 × 3 = 544 N
1.7 Accelerating downward at −3 m·s−2:
FN = mg − m(−3) = 80 × 9.8 + 80 × 3 = 1024 N
1.8 Free fall:
FN = 0
Answer: The scale reading is zero. This is the case of weightlessness.