3 Questions about Rope tension equation
1. The picture below shows three blocks, namely A, B and C which are located on a smooth horizontal plane. If mass A = 1 kg, mass B = 2 kg and mass C = 2 kg and F = 10 N, then determine the ratio of the tension in the rope between A and B to the tension in the rope between B and C.
Known:
The mass of A (mA) = 1 kg
Mass B (mB) = 2 kg
The mass of C (mC) = 2 kg
Tensile force (F) = 10 N
Wanted: TAB : TBC
Solution:
Calculate the acceleration of the system using Newton’s Second Law formula:
ΣF = m a
F = (mA + mB + mC) a
10 = (1 + 2 + 2) a
10 = 5 a
a = 10 / 5
a = 2 m/s2
Use the rope tension formula to calculate TAB
ΣF = m a
TAB = mA a = 1 (2) = 2 Newton
Use the rope tension formula to calculate TBC
ΣF = m a
TBC = (mA + mB) a = (1 + 2) (2) = (3)(2) = 6 Newton
2. Object A with a mass of 6 kg and object B with a mass of 3 kg are connected by a rope as shown. If the coefficient of friction is 0.3 and g = 10 m/s2, determine the acceleration of the object and the tension in the ropes of each block.
Known:
The mass of object A (mA) = 6 kg
The mass of object B (mB) = 3 kg
Coefficient of friction of block A (µk) = 0.3
Acceleration due to gravity (g) = 10 m/s2
The weight of block A (wA) = mA g = (6)(10) = 60 N
Normal force on block A (NA) = wA = 60 N
The weight of block B (wB) = mB g = (3)(10) = 30 N
Wanted: The acceleration of the system (a) and the tension in the rope (T)
Solution:
Calculate the kinetic frictional force i.e. the frictional force when block A moves:
Fk = µk NA = (0,3)(60) = 18 Newton
Calculate the acceleration of the system (a):
ΣF = m a
wB – Fk = (mA + mB) a
30 – 18 = (6 + 3) a
12 = 9 a
a = 12 / 9 = 1,3 m/s2
Calculate the tension in the string on block A (TA):
ΣF = m a
TA – Fk = (mA) a
TA – 18 = (6)(1,3)
TA – 18 = 7,8
TA = 7,8 + 18 = 25,8 Newton
Calculate the tension in the rope on beam B (TB):
ΣF = m a
wB – TB = mB (a)
30 – TB = 3 (1,3)
30 – TB = 3,9
TB = 30 – 3,9
TB = 26,1 Newton
3. Two objects A and B with masses of 5 kg and 3 kg are connected by a frictionless pulley. The force P is applied to the pulley in an upward direction. If both blocks are initially at rest on the floor, what is the acceleration of block A, if the magnitude of P is 60 N?
Determine also the tension in the rope on blocks A and B.
Known:
Acceleration due to gravity (g) = 10 m/s2
The mass of A (mA) = 5 kg
The weight of block A (wA) = mA g = (5)(10) = 50 Newton
Mass B (mB) = 3 kg
The weight of block B (wB) = mB g = (3)(10) = 30 Newton
Force P = 60 N
Wanted: Acceleration of the system of beams A and B (a) and the tension in the rope on beam A (TA) and beam B (TB)
Solution:
Calculate the acceleration of the system using Newton’s Second Law formula.
ΣF = m a
wA – wB = (mA + mB) a
50 – 30 = (5 + 3) a
20 = 8 a
a = 20 / 8
a = 2,5 m/s2
Use the tension force formula to calculate the tension in the rope
The tension in the rope on block A:
ΣF = m a
wA – TA = mA a
50 – TA = 5 (2,5)
50 – TA = 12,5
TA = 50 – 12,5 = 37,5 Newton
The tension in the rope on block B:
ΣF = m a
TB – wB = mB a
TB – 30 = 3 (2,5)
TB – 30 = 7,5
TB = 7,5 + 30 = 37,5 Newton
