Normal na puwersa – mga problema at solusyon

Solved problems in Newton’s laws of motion – Normal force 

1. An object resting on a table, shown in the figure below. Mass of the object is 1 kg. Pagpapabilis ng grabidad ay 9.8 m/s2. Determine the normal force exerted on the object by the table.

Normal-force-–-problems-and-solutions-1-1

Kilala:

Mass (m) = 1 kg

Akselerasyon ng grabidad (g) = 9.8 m/s2

Timbang (w) = mg = (1 kg)(9.8 m/s2) = 9.8 kg m/s2 = 9.8 Newton

Wanted: normal force (N)

solusyon:

Normal force – problems and solutions 2

The object is at rest on the table, so the net force on the object is zero (Newton’s first or second law). The weight of the object acts vertically downward, toward the center of the Earth. There must be another force on the object to balance the puwersang gravitational. Object resting on the table, so that the table exerts this upward force. The force exerted by the table is often called a normal force (N). Normal means perpendicular.

Choose the upward direction as the positive y-direction. The net force on the object is :

ΣFy = 0

N – w = 0

N = w

N = m g

N = 9.8 Newton

The normal force on the object, exerted by the table is 9.8 N upward.

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2. Two objects resting on a table. Masa of object 1 (m1) = 1 kg, mass of object 2 (m2) = 2 kg, acceleration due to gravity (g) =9.8 m/s2. Determine the magnitude and direction of the normal force exerted by m2 on the m1 and the normal force exerted by the table on the m2.

Normal force – problems and solutions 3

Solusyon

Normal force – problems and solutions 4

Kilala:

Masa ng bagay 1 (m1) = 1 kilo

Masa ng bagay 2 (m2) = 2 kilo

Akselerasyon ng grabidad (g) = 9.8 m/s2

timbang of object 1 (w1) = m1 g = (1)(9.8 m/s2) = 9.8 kg m/s2 = 9.8 Newton

Weight of object 2 (w2) = m2 g = (2)(9.8 m/s2) = 19.6 kg m/s2 = 19.6 Newton

Hinahanap: N1 at N2

solusyon:

(a) Normal force exerted by m2 to the m1 (N1)

N1 = w1 = 9.8 Newton

Direction of N1 is upward.

(b) Normal force exerted by the table on the m2 (N2)

N2 = w1 +w2 = 9.8 Newton + 19.6 Newton = 29.4 Newton

Direction of N2 is upward.

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3. An object resting on the table. Mass of the object is 2 kg, acceleration due to gravity is 9.8 m/s2. Magnitude of the force F is 10 Newton. Find the magnitude and direction of the normal force exerted by the table on the object.

Normal force – problems and solutions 5

Solusyon

Normal force – problems and solutions 6

Kilala:

Mass of the object (m) = 2 kg

Akselerasyon dahil sa grabidad (g) = 9.8 m/s2

Timbang (w) = mg = (2 kg)(9.8 m/s2) = 19.6 kg m/s2 = 19.6 Newton

Force F (F) = 10 Newton

Karaniwan : magnitude and direction of the normal force (N)

solusyon:

direction of the normal force is upward.

Magnitude of the normal force :

ΣF=0

N – F – w = 0

N = F + w

N = 10 Newton + 20 Newton

N = 30 Newton

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4. An object resting on a table. Object’s mass is 1 kg, acceleration due to gravity is 9,8 m/s2, force F1 is 10 N and force F2 is 20 N. Determine magnitude and direction of the normal force exerted by the table on the object. g = 9.8 m/s2

Normal force – problems and solutions 7

Solusyon

Normal force – problems and solutions 8

Kilala:

Mass (m) = 1 kg

Akselerasyon ng grabidad (g) = 9.8 m/s2

Timbang (w) = mg = (1 kg)(9.8 m/s2) = 9.8 kg m/s2 = 9.8 Newton

F1 = 10 Newton

F2 = 20 Newton

Hinahanap: magnitude and direction of the normal force (N)

solusyon:

Direction of the normal force is upward.

Magnitude of the normal force :

ΣF=0

N – F2 – w + F1 = 0

N = F2 + w – F1

N = 20 Newton + 9.8 Newton – 10 Newton

N = 19.8 Newton

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5. Object’s mass (m) = 2 kg, acceleration of gravity (g) = 9.8 m/s2, angle = 30o. Find magnitude and direction of the normal force exerted on the object.

Normal force – problems and solutions 9

solusyon:

Normal force – problems and solutions 10

w is weight, wx is horizontal component of the weight, wy is a vertical component of the weight, N is the normal force.

Kilala:

mass (m) = 2 kg

acceleration of gravity (g) = 9.8 m/s2

timbang (w) = mg = (2 kg)(9.8 m/s2) = 19.6 kg m/s2 = 19.6 Newton

wx = w sin 60o = (19.6 N)(0.5)√3= 9.8√3 Newton

wy = w cos 60 = (19.6 N)(0.5) = 9.8 Newton

Wanted: normal na puwersa (N)

solusyon:

ΣF=0

N – wy = 0

N = wy

N = 9.8 Newton

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  2. Normal na pwersa
  3. Ikalawang batas ng galaw ni Newton
  4. Pwersa ng friction
  5. Paggalaw sa pahalang na ibabaw nang walang puwersa ng alitan
  6. Ang galaw ng dalawang katawan na may parehong acceleration sa magaspang na pahalang na ibabaw na may puwersa ng friction
  7. Paggalaw sa inclined plane nang walang friction force
  8. Paggalaw sa magaspang na hilig na eroplano na may puwersa ng friction
  9. Paggalaw sa isang elevator
  10. Ang paggalaw ng mga katawan ay konektado sa pamamagitan ng mga kordon at pulley
  11. Dalawang katawan na may parehong magnitude ng acceleration
  12. Pag-ikot ng patag na kurba – dinamika ng pabilog na galaw
  13. Pag-ikot sa isang naka-bankong kurba – dinamika ng pabilog na galaw
  14. Pare-parehong galaw sa isang pahalang na bilog
  15. Puwersang sentripetal sa pare-parehong pabilog na galaw

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