Inclined plane – problems and solutions

Inclined plane – problems and solutions

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?

Inclined plane – problems and solutions 1

Solution

Known :Inclined plane – problems and solutions 2

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

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 :Inclined plane – problems and solutions 3

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 Fwx + fs. Inclined plane – problems and solutions 3

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

Normal force :

N = wy = 40√2

Force of static friction :

fs = µs N = (0.5)(40√2) = 20√2

Magnitude of force F :

Fwx + fs

F ≥ 40√2 + 20√2

F ≥ 60√2 Newton

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 :Inclined plane – problems and solutions 5

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

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Normal force (N) = wy = w cos 37o = (80)(0.8) = 64 Newton

Wanted : Magnitude of force F

Solution :Inclined plane – problems and solutions 6

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

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 kgInclined plane – problems and solutions 7

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 :Inclined plane – problems and solutions 8

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 :Inclined plane – problems and solutions 5

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 :

Inclined plane – problems and solutions 6Sin 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

  1. What is an inclined plane, and how does it make work easier?
    • Answer: An inclined plane is a flat surface set at an angle (other than a right angle) against a horizontal surface. It makes work easier by allowing a force to be applied over a longer distance to overcome a resistive force, often reducing the magnitude of the force needed to lift or lower an object.
  2. How does the angle of inclination affect the force required to move an object up the plane?
    • Answer: The steeper the angle of inclination, the greater the force required to move an object up the plane. As the angle decreases, the required force diminishes but is applied over a longer distance.
  3. Why is it easier to push a heavy box up a ramp than to lift it straight up?
    • Answer: Pushing a box up a ramp (an inclined plane) divides the weight of the box over a greater distance, reducing the force required at any given moment. Lifting it straight up requires one to counteract the full weight of the box directly, necessitating a much larger force.
  4. How does the length of the inclined plane relate to the force required to move an object up the plane?
    • Answer: The longer the inclined plane (for a fixed height), the less force is required to move an object up the plane. However, the force must be applied over a longer distance.
  5. If friction is introduced on the inclined plane, how does it affect the motion of an object on it?
    • Answer: Friction opposes the motion of the object. If an object is being pushed up the inclined plane, friction acts downward, increasing the force needed to move the object. If an object is moving down the plane, friction slows its descent.
  6. Why do workers use inclined planes to load heavy items onto trucks?
    • Answer: Using an inclined plane (ramp) allows workers to use a smaller force over a longer distance to load heavy items, making the task more manageable and less strenuous than lifting the items vertically.
  7. How does the weight of an object relate to its tendency to slide down an inclined plane?
    • Answer: The component of the object’s weight acting parallel to the surface of the inclined plane causes it to slide. The heavier the object, the greater this component of weight, and thus, the greater the tendency for the object to slide down.
  8. What role does the normal force play on an inclined plane?
    • Answer: The normal force acts perpendicular to the surface of the inclined plane. It’s responsible for the amount of frictional force between the object and the plane since friction is usually proportional to the normal force.
  9. Why might sand or grit be placed on an inclined plane during icy conditions?
    • Answer: Sand or grit increases the frictional force between the inclined plane and objects (like shoes or tires) in contact with it. This added friction helps prevent slipping during icy or wet conditions.
  10. If a ball is rolled up an inclined plane and then rolls back down, will it reach its original speed by the time it returns to its starting point, ignoring air resistance and friction?
  • Answer: Yes, in the absence of air resistance and friction, the ball’s potential energy at its highest point will convert fully back to kinetic energy as it descends, allowing it to reach its original speed by the time it returns to its starting point due to the conservation of energy.
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