1. Object A with a mass of 6-kg and object B with a mass of 4-kg connected by a cord and pulled by a force of F = 60 N, as shown in the figure below. The coefficient of kinetic friction between the floor and both objects is 0.5 (tan θ = ¾). Acceleration due to gravity is 10 m/s2. What is the magnitude of the tension force?
Known :
Mass of object A (mA) = 6 kg
Mass of object B (mB) = 4 kg
Force (F) = 60 Newton
The coefficient of kinetic friction between object and floor (μk) = 0.5
Acceleration due to gravity (g) = 10 m/s2
Tan θ = 3/4
Wanted : Tension force (T)
Solution :
Horizontal component of force F :
Fx = F cos θ
Fx = (60)(4/5) = (4)(12) = 48 N
Vertical component of force F :
Fy = F sin θ
Fy = (60)(3/5) = (3)(12) = 36 N
The normal force on object A :
NA = wA = mA g = (6)(10) = 60 N
The normal force on object B :
NB + Fy – wB = 0
NB + Fy = wB
NB = wB – Fy = mB g – Fy = (4)(10) – 36 = 40 – 36 = 4 N
The force of kinetic friction between object A and floor :
fkA = μk NA = (0.5)(60) = 30 N
The force of kinetic friction between object B and floor :
fkB = μk NB = (0.5)(4) = 2 N
Calculate the acceleration of both objects :
ΣF = m a
Fx – T + T – fkB – fkA = (mA + mB) a
Fx – fkB – fkA = (mA + mB) a
48 – 2 – 30 = (6 + 4) a
16 = 10 a
a = 16/10
a = 1.6 m/s2
Calculate the tension force :
Object A :
ΣF = m a
TA – fkA = mA a
TA – 30 = (6)(1.6)
TA – 30 = 9.6
TA = 9.6 + 30 = 39.6 N
Object B :
ΣF = m a
Fx – fkB – TB = mB a
48 – 2 – TB = (4)(1.6)
46 – TB = 6.4
46 – 6.4 = TB
TB = 39.6 N
[irp]
2. If the coefficient of kinetic friction between both blocks and floor is 0.2 , what is the acceleration of both objects ? (cos 37o = 0,8, sin 37o = 0,6)
Known :
Mass of object A (mA) = 4 kg
Mass of object B (mB) = 2 kg
Force (F) = 30 Newton
The coefficient of kinetic friction between object and floor (μk) = 0.2
Acceleration due to gravity (g) = 10 m/s2
cos 37o = 0.8
sin 37o = 0.6
Wanted : Acceleration of both objects
Solution :
The horizontal component of force F :
Fx = F cos θ
Fx = (30)(0.8) = 24 N
The vertical component of force F :
Fy = F sin θ
Fy = (30)(0.6) = 18 N
The normal force on object A :
NA = wA = mA g = (4)(10) = 40 N
The normal force on object B :
NB + Fy – wB = 0
NB + Fy = wB
NB = wB – Fy = mB g – Fy = (2)(10) – 18 = 20 – 18 = 2 N
The force of kinetic friction between object A and floor :
fkA = μk NA = (0.2)(40) = 8 N
The force of kinetic friction between object B and floor :
fkB = μk NB = (0.2)(2) = 0.4 N
Acceleration of both objects :
ΣF = m a
Fx – fkB – fkA = (mA + mB) a
24 – 0.4 – 8 = (4 + 2) a
15.6 = 6 a
a = 15.6 / 6
a = 2.6 m/s2
[irp]
3. Two objects connected by a cord over a pulley as shown in figure below. Mass of object A = mA, mass f object B = mB and the acceleration of block B is a. Acceleration due to gravity is g. What is the tension force on block B.
Solution :
The horizontal surface is smooth so there is no friction. The only force that accelerates system is the weight of block B.
System’s acceleration :
The tension force (T) :
Substitute mA in equation 1 with mA in equation 2.
[irp]
