Налуу хавтгай - асуудал ба шийдэл

Налуу хавтгай - асуудал ба шийдэл

1. Масс of block is 5 kg, таталцлын улмаас үүссэн хурдатгал = 9.8 м/с², tan 37 ^o = 3/4. What is the acceleration of the block?

шийдэл

Мэдэгдэж байгаа:

масс (м) = 5 кг

таталцлын хүчнээс үүдэлтэй хурдатгал (g) = 10 м/с²

жин (w) = мг = (5)(10) = 50 N

х = 3 байна

Өндөр налуу хавтгай (y) = 4

The length of inclined plane surface (l) = 5

Wanted : хурдатгал of block (a) ?

Шийдэл:

The block is accelerated by force of w_x. Inclined plane is smooth so there is no үрэлтийн хүч.

w_x = w sin θ = (w)(y / z)

w_x = (50)(4/5)

w_x = 40 Ньютон

Acceleration of block (a) ?

ΣF = ma

w_x = ма

40 = (5) а

а = 40 / 5

a = 8 м/с²

2. A block initially at rest, then accelerated upward parallel with the inclined plane surface. Block’s mass = 8 kg, coefficient of static friction µ_s = 0.5 and angle θ = 45^o. What is the magnitude of force F so that the block is accelerated.

Мэдэгдэж байгаа:

Coefficient of static friction (µ_s) = 0.5

Өнцөг (θ) = 45^o

Таталцлын хүчнээс үүдэлтэй хурдатгал (g) = 10 м/с²

Блокийн масс (м) = 8 кг

Block weight (w) = m g = (8 kg)(10 m/s²) = 80 кг м/с² = 80 Ньютон

Wanted : Magnitude of force F

Шийдэл:

Block start to accelerated if F ≥ w_x + f_s.

Horizontal component of weight :

w_x = w sin θ = (80)(sin 45) = (80)(0.5√2) = 40√2

Vertical component of weight :

w_y = w cos θ = (80)(cos 45) = (80)(0.5√2) = 40√2

Ердийн хүч :

N = w_y = 40√2

Force of static friction :

f_s = µ_s N = (0.5)(40√2) = 20√2

Magnitude of force F :

F ≥ w_x + f_s

F ≥ 40√2 + 20√2

F ≥ 60√2 Ньютон

3. Mass of block is 8 kg. The minimal force required to hold the блок so that the block not slides down is …(нүгэл 37^o = 0.6, cos 37^o = 0.8, g = 10 мс^-2, µ_k = 0.1)

Мэдэгдэж байгаа:

Блокны масс (м) = 8 кг

Таталцлын хүчнээс үүдэлтэй хурдатгал (g) = 10 м/с²

Блокийн жин (w) = mg = (8)(10) = 80 Ньютон

Нүгэл 37^o = 0.6

Кос 37^o = 0.8

Coefficient of kinetic friction (µ_k) = 0.1

Хэвийн хүч (N) = w_y = w cos 37^o = (80)(0.8) = 64 Ньютон

Хүссэн: Хүчний хэмжээ F

Шийдэл:

The block is not slides down if F = w_x

Horizontal component of weight parallel with inclined plane surface :

w_x = w sin θ = (80)(sin 37) = (80)(0.6) = 48

f_k = µ_k N = (0.1)(64) = 6.4

The minimum of force F so that the block is not slides down :

F + f_k - w_x = 0

F = w_x - f_k = 48 – 6.4 = 41.6 Ньютон

4. A 5-kg block is pulled along rough inclined plane by a force of 71 N (g = 10 m.s^-2, нүгэл 37^o = 0.6, cos 37^o = 0.8). If the coefficient of friction force between block and inclined plane is 0.4, what is the acceleration of the block.

Мэдэгдэж байгаа:

Object’s mass (m) = 5 kg

Таталцлын хүчнээс үүдэлтэй хурдатгал (g) = 10 м/с²

weight of block (w) = m g = (5)(10) = 50 Newton

Force F = 71 Newton

Нүгэл 37^o = 0.6

Кос 37^o = 0.8

Coefficient of friction = 0.4

Wanted : Acceleration of the block (a)

Шийдэл:

Horizontal component of weight :

w_x = w sin θ = (50)(sin 37) = (50)(0.6) = 30

Vertical component of weight :

w_y = w cos θ = (50)(cos 37) = (50)(0.8) = 40

Normal force :

N = w_y = 40

Force of kinetic friction :

f_k = µ_k N = (0.4)(40) = 16

Цэвэр хүч:

∑F = F – w_x - f_k = 71 – 30 – 16 = 25 Ньютон

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.

Мэдэгдэж байгаа:

weight of box (w) = 500 Newton

height of truck (h) = 1.5 meters

length of inclined plane (l) = 3 meters

Хүссэн: Magnitude of force to moving the box (F)

Шийдэл:

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 (w_x).

w_x = w sin θ = (500 N)(0.5) = 250 N

1. What is an inclined plane, and how does it make work easier?
   • хариулт: 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?
   • хариулт: 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?
   • хариулт: 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?
   • хариулт: 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?
   • хариулт: 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?
   • хариулт: 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?
   • хариулт: 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?
   • хариулт: 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?
   • хариулт: 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?

• хариулт: 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.