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/s^{2}. Determine the normal force exerted on the object by the table.

__Known :__

Mass (m) = 1 kg

Acceleration of gravity (g) = 9.8 m/s^{2}

Weight (w) = m g = (1 kg)(9.8 m/s^{2}) = 9.8 kg m/s^{2} = 9.8 Newton

__Wanted:__ normal force (N)

__Solution :__

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.

[irp]

2. Two objects resting on a table. Mass of object 1 (m_{1}) = 1 kg, mass of object 2 (m_{2}) = 2 kg, acceleration due to gravity (g) =9.8 m/s^{2}. Determine the magnitude and direction of the normal force exerted by m_{2} on the m_{1} and the normal force exerted by the table on the m_{2}.

Solution

__Known :__

Mass of the object 1 (m_{1}) = 1 kg

Mass of the object 2 (m_{2}) = 2 kg

Acceleration of gravity (g) = 9.8 m/s^{2}

Weight of object 1 (w_{1}) = m_{1} g = (1)(9.8 m/s^{2}) = 9.8 kg m/s^{2} = 9.8 Newton

Weight of object 2 (w_{2}) = m_{2} g = (2)(9.8 m/s^{2}) = 19.6 kg m/s^{2} = 19.6 Newton

__Wanted :__ N_{1} and N_{2}

__Solution :__

(a) Normal force exerted by m_{2} to the m_{1} (N_{1})

N_{1} = w_{1} = 9.8 Newton

Direction of N_{1} is upward.

(b) Normal force exerted by the table on the m_{2} (N_{2})

N_{2} = w_{1} + w_{2} = 9.8 Newton + 19.6 Newton = 29.4 Newton

Direction of N_{2} is upward.

[irp]

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

Solution

__Known :__

Mass of the object (m) = 2 kg

Acceleration due to gravity (g) = 9.8 m/s^{2}

Weight (w) = m g = (2 kg)(9.8 m/s^{2}) = 19.6 kg m/s^{2} = 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

[irp]

4. An object resting on a table. Object’s mass is 1 kg, acceleration due to gravity is 9,8 m/s^{2}, force F_{1} is 10 N and force F_{2} is 20 N. Determine magnitude and direction of the normal force exerted by the table on the object. g = 9.8 m/s^{2}

__Solution__

__Known :__

Mass (m) = 1 kg

Acceleration of gravity (g) = 9.8 m/s^{2}

Weight (w) = m g = (1 kg)(9.8 m/s^{2}) = 9.8 kg m/s^{2} = 9.8 Newton

F_{1} = 10 Newton

F_{2} = 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 – F_{2} – w + F_{1} = 0

N = F_{2} + w – F_{1}

N = 20 Newton + 9.8 Newton – 10 Newton

N = 19.8 Newton

[irp]

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

__Solution :__

w is weight, w_{x} is horizontal component of the weight, w_{y} is a vertical component of the weight, N is the normal force.

__Known :__

mass (m) = 2 kg

acceleration of gravity (g) = 9.8 m/s^{2}

weight (w) = m g = (2 kg)(9.8 m/s^{2}) = 19.6 kg m/s^{2} = 19.6 Newton

w_{x} = w sin 60^{o} = (19.6 N)(0.5)√3= 9.8√3 Newton

w_{y} = w cos 60 = (19.6 N)(0.5) = 9.8 Newton

__Wanted:__ normal force (N)

__Solution :__

∑F = 0

N – w_{y} = 0

N = w_{y}

N = 9.8 Newton

[irp]

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