1. Mass of block is 5 kg, acceleration due to gravity = 9.8 m/s2, tan 37 o = 3/4. What is the acceleration of the block?
Solution
Known :
mass (m) = 5 kg
acceleration due to gravity (g) = 10 m/s2
weight (w) = m g = (5)(10) = 50 N
x = 3
The height of inclined plane (y) = 4
The length of inclined plane surface (l) = 5
Wanted : acceleration of block (a) ?
Solution :
The block is accelerated by force of wx. Inclined plane is smooth so there is no friction force.
wx = w sin θ = (w)(y / z)
wx = (50)(4/5)
wx = 40 Newton
Acceleration of block (a) ?
ΣF = m a
wx = m a
40 = (5) a
a = 40 / 5
a = 8 m/s2
[irp]
2. A block initially at rest, then accelerated upward parallel with the inclined plane surface. Block’s mass = 8 kg, coefficient of static friction µs = 0.5 and angle θ = 45o. What is the magnitude of force F so that the block is accelerated.
Known :
Coefficient of static friction (µs) = 0.5
Angle (θ) = 45o
Acceleration due to gravity (g) = 10 m/s2
Block’s mass (m) = 8 kg
Block weight (w) = m g = (8 kg)(10 m/s2) = 80 kg m/s2 = 80 Newton
Wanted : Magnitude of force F
Solution :
Block start to accelerated if F ≥ wx + fs.
Horizontal component of weight :
wx = w sin θ = (80)(sin 45) = (80)(0.5√2) = 40√2
Vertical component of weight :
wy = w cos θ = (80)(cos 45) = (80)(0.5√2) = 40√2
N = wy = 40√2
Force of static friction :
fs = µs N = (0.5)(40√2) = 20√2
Magnitude of force F :
F ≥ wx + fs
F ≥ 40√2 + 20√2
F ≥ 60√2 Newton
[irp]
3. Mass of block is 8 kg. The minimal force required to hold the block so that the block not slides down is …(sin 37o = 0.6, cos 37o = 0.8, g = 10 m.s-2, µk = 0.1)
Known :
Mass of block (m) = 8 kg
Acceleration due to gravity (g) = 10 m/s2
Block’s weight (w) = m g = (8)(10) = 80 Newton
Sin 37o = 0.6
Cos 37o = 0.8
Coefficient of kinetic friction (µk) = 0.1
Normal force (N) = wy = w cos 37o = (80)(0.8) = 64 Newton
Wanted : Magnitude of force F
Solution :
The block is not slides down if F = wx
Horizontal component of weight parallel with inclined plane surface :
wx = w sin θ = (80)(sin 37) = (80)(0.6) = 48
fk = µk N = (0.1)(64) = 6.4
The minimum of force F so that the block is not slides down :
F + fk – wx = 0
F = wx – fk = 48 – 6.4 = 41.6 Newton
[irp]
4. A 5-kg block is pulled along rough inclined plane by a force of 71 N (g = 10 m.s-2, sin 37o = 0.6, cos 37o = 0.8). If the coefficient of friction force between block and inclined plane is 0.4, what is the acceleration of the block.
Known :
Object’s mass (m) = 5 kg
Acceleration due to gravity (g) = 10 m/s2
weight of block (w) = m g = (5)(10) = 50 Newton
Force F = 71 Newton
Sin 37o = 0.6
Cos 37o = 0.8
Coefficient of friction = 0.4
Wanted : Acceleration of the block (a)
Solution :
Horizontal component of weight :
wx = w sin θ = (50)(sin 37) = (50)(0.6) = 30
Vertical component of weight :
wy = w cos θ = (50)(cos 37) = (50)(0.8) = 40
Normal force :
N = wy = 40
Force of kinetic friction :
fk = µk N = (0.4)(40) = 16
Net force :
∑F = F – wx – fk = 71 – 30 – 16 = 25 Newton
Acceleration of the object :
a = ∑F / m = 25 / 5 = 5 m/s2
5. A 500-N box raised to truck through inclined plane, as shown in figure below. If height of truck is 1.5 meters, determine the magnitude of force to moving the box.
Known :
weight of box (w) = 500 Newton
height of truck (h) = 1.5 meters
length of inclined plane (l) = 3 meters
Wanted : Magnitude of force to moving the box (F)
Solution :
Sin θ = opp / hyp = 1.5 meters / 3 meters = 1.5 / 3 = 0.5
Force (F) required to moving the box, same as the horizontal component of weight which is parallel to inclined plane (wx).
wx = w sin θ = (500 N)(0.5) = 250 N