Normal force – problems and solutions

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. Acceleration of gravity is 9.8 m/s2. Determine the normal force exerted on the object by the table.

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

Known :

Mass (m) = 1 kg

Acceleration of gravity (g) = 9.8 m/s2

Weight (w) = m g = (1 kg)(9.8 m/s2) = 9.8 kg m/s2 = 9.8 Newton

Wanted: normal force (N)

Solution :

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 gravitational force. 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 :

∑Fy = 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. Mass 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

Solution

Normal force – problems and solutions 4

Known :

Mass of the object 1 (m1) = 1 kg

Mass of the object 2 (m2) = 2 kg

Acceleration of gravity (g) = 9.8 m/s2

Weight 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

Wanted : N1 and N2

Solution :

(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

Solution

Normal force – problems and solutions 6

Known :

Mass of the object (m) = 2 kg

Acceleration due to gravity (g) = 9.8 m/s2

Weight (w) = m g = (2 kg)(9.8 m/s2) = 19.6 kg m/s2 = 19.6 Newton

Force F (F) = 10 Newton

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

Solution :

direction of the normal force is upward.

Magnitude of the normal force :

∑F = 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

Solution

Normal force – problems and solutions 8

Known :

Mass (m) = 1 kg

Acceleration of gravity (g) = 9.8 m/s2

Weight (w) = m g = (1 kg)(9.8 m/s2) = 9.8 kg m/s2 = 9.8 Newton

F1 = 10 Newton

F2 = 20 Newton

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

Solution :

Direction of the normal force is upward.

Magnitude of the normal force :

∑F = 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

Solution :

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.

Known :

mass (m) = 2 kg

acceleration of gravity (g) = 9.8 m/s2

weight (w) = m g = (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 force (N)

Solution :

∑F = 0

N – wy = 0

N = wy

N = 9.8 Newton

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  1. Mass and weight
  2. Normal force
  3. Newton’s second law of motion
  4. Friction force
  5. Motion on the horizontal surface without friction force
  6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
  7. Motion on the inclined plane without friction force
  8. Motion on the rough inclined plane with the friction force
  9. Motion in an elevator
  10. The motion of bodies connected by cord and pulley
  11. Two bodies with the same magnitude of accelerations
  12. Rounding a flat curve – dynamics of circular motion
  13. Rounding a banked curve – dynamics of circular motion
  14. Uniform motion in a horizontal circle
  15. Centripetal force in uniform circular motion

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