Can normal force be negative

Slope driving force / normal force

A skater is also subject to the slope driving force

To understand the slope driving force, a body on a ramp is considered:

Normal force and weight force

 

Which forces act on the body?

The body is pulled down by gravity. Depending on the mass of the body under consideration, the weight, directed vertically downwards, acts with $ G = mg $, which acts in the center of gravity of the body.

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The weight force is always directed vertically downwards and acts in the center of gravity of the body.

Because of the ramp, however, the body remains on the ramp. The ramp prevents the body from accelerating towards the center of the earth. So the force that the ramp exerts on the body also acts. This force is called normal force $ F_N $ or $ N $.

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The normal force is always perpendicular to the contact surface.

These are the two forces acting on the body. If we next lay the $ x $ axis in the direction of the movement of the body, we can clearly see that the weight has a component in the $ x $ direction and a component in the $ y $ direction. The component of the weight in the direction of the $ x $ axis causes the body to move downwards. The larger the angle $ \ alpha $, the larger the $ x $ component of the weight and the faster the body moves.

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A so-called breakdown of forces must be carried out. The weight, which is directed vertically downwards, is divided into a component parallel to the contact surface ($ x $ axis) and a component perpendicular to the contact surface ($ y $ axis).

For this it is necessary to set the $ x, y $ coordinate system in such a way that the x-axis runs in the direction of the movement, i.e. inclined by $ \ alpha $ to the horizontal:

Decomposition of the weight force

The $ x $ component of the weight force (i.e. the part that is responsible for the movement) is also referred to as the slope driving force $ F_ {slope} $. The normal force $ F_N $ is equal to the $ y $ component of the weight force. The normal force replaces the plane on which the body is located and is always drawn in perpendicular (at a 90 ° angle) to the plane.

It follows from this:

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$ F_ {Hang} = G \ cdot \ sin (\ alpha) $ (parallel to the contact surface)

$ F_N = G \ cos (\ alpha) $ (perpendicular to the contact surface)

The slope driving force points exactly in the direction of the $ x $ component of the weight force, while the normal force points in the opposite direction to the $ y $ component.

The slope driving force increases with increasing mass and increasing angle. If it is a horizontal ramp, the slope driving force is $ F = _ {slope} = 0 $, since the angle $ \ alpha = 0 ° $ and the normal force $ N = G \ cos \ alpha = 9.81 N $ then assumes its maximum value, ie normal force and weight force are then identical. If it is a vertical ramp with $ \ alpha = 90 ° $, the normal force disappears and the slope driving force takes its maximum value, i.e. it is equal to the weight force (body moves in free fall).

If a mass is accelerated by the slope driving force, the equation of motion can be written as:

$ F = ma $

Where $ F $ represents the slope driving force with $ F = G \ cdot \ sin (\ alpha) = mg \ cdot \ sin (\ alpha) $:

$ mg \ cdot \ sin (\ alpha) = ma $ |: m

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$ a = g \ cdot \ sin (\ alpha) $

No acceleration has to be introduced for the normal force, since it does not move the body. The nominal force is necessary to describe the friction and adhesion of a body. With the help of the normal force, the frictional force can be determined.

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The slope driving force is the $ x $ component of the weight force. The normal force is the $ y $ component of the weight force. In order for this relationship to exist, the $ x $ axis must be placed in the direction of the movement. Then the weight (which is always directed vertically downwards) is broken down into its $ x $ and $ y $ components.

Application example: slope driving force and normal force

Example: slope driving force and normal force
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The above box is given with $ m = 20 kg $ on the inclined plane with the angle $ \ alpha $ to the horizontal. Calculate the slope driving force and normal force for $ \ alpha = 25 ° $ and $ \ alpha = 60 ° $. Also determine the acceleration for both angles!

First, the box is cut free:

After that we put the $ x $ axis in the direction of movement:

x-axis in the direction of movement

We can next determine the slope driving force and the normal force by dividing the weight force into its $ x $ and $ y $ components. The $ x $ component of the weight force is called the slope driving force:

$ F_ {hang} = G \ cdot \ sin (\ alpha) $


The normal force $ F_N $ is equal to the $ y $ component of the weight:

$ F_N = G \ cdot \ cos (\ alpha) $

The forces in the $ y $ direction cancel each other out because the $ y $ component of the weight force points in the direction of the negative $ y $ axis and the normal force in the direction of the positive $ y $ axis (forces are directed in opposite directions) and both have the same amount. There is therefore no movement in the $ y $ direction.

If a force is broken down into its components, these replace the force:

Components of weight

The slope driving force is the $ x $ component of the weight force and is:

$ F_ {slope} = G \ cdot \ sin (25 °) = 20 kg \ cdot 9.81 \ frac {m} {s ^ 2} \ cdot \ sin (25 °) = 82.92 \ frac {kg \ ; m} {s ^ 2} = 82.92N $

$ F_ {slope} = G \ cdot \ sin (60 °) = 20 kg \ cdot 9.81 \ frac {m} {s ^ 2} \ cdot \ sin (60 °) = 82.92 \ frac {kg \ ; m} {s ^ 2} = 169.91N $

The normal force is:

$ F_N = G \ cdot \ cos (25 °) = 20 kg \ cdot 9.81 \ frac {m} {s ^ 2} \ cdot \ cos (25 °) = 82.92 \ frac {kg \; m} {s ^ 2} = N $ 177.82

$ F_N = G \ cdot \ cos (60 °) = 20 kg \ cdot 9.81 \ frac {m} {s ^ 2} \ cdot \ cos (60 °) = 82.92 \ frac {kg \; m} {s ^ 2} = 98.1 N $

The larger the angle $ \ alpha $, the greater the slope driving force and the lower the normal force. At an angle of 90 °, the slope driving force is maximum and the normal force is zero. The body is then in free fall and the entire weight force goes into the movement of the body.

We can also determine the acceleration of the body. Newton's basic law reads:

$ F_x = ma_x $

$ F_y = ma_y $

Since there is no movement in the $ y $ direction, $ a_y = 0 $. So $ F_y = 0 $ applies, where $ F_y $ represents the sum of all forces in the $ y $ direction. This equation also gives us the relationship that the normal force is equal to the $ y $ component of the weight:

$ F_y = -G \ cdot \ cos (\ alpha) + F_N $ | with $ F_y = 0 $

$ -G \ cdot \ cos (\ alpha) + F_N = 0 $ | Solve for $ F_N $

$ F_N = G \ cdot \ cos (\ alpha) $

From the equation $ F_x = ma_x $ we can determine the acceleration $ a = a_x $:

$ F_x = G \ cdot \ sin (\ alpha) $ | Insert

Solve $ G \ cdot \ sin (\ alpha) = m a_x $ | for $ a_x $

$ a_x = \ frac {G \ cdot \ sin (\ alpha)} {m} = Shorten \ frac {m \ cdot g \ cdot \ sin (25 °)} {m} $ | $ m $

$ a_x = g \ cdot \ sin (\ alpha) $


In this case, the acceleration is only dependent on the angle $ \ alpha $. The acceleration due to gravity is constant:

$ a_x = 9.81 \ frac {m} {s ^ 2} \ cdot \ sin (25 °) = 4.15 \ frac {m} {s ^ 2} $

$ a_x = 9.81 \ frac {m} {s ^ 2} \ cdot \ sin (60 °) = 8.50 \ frac {m} {s ^ 2} $

Application example: box and tipping height

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The above box with $ m = 80kg $, which is on a frictionless surface, is given. A force $ F = 240 N $ acts on the box.

a) Determine the acceleration of the box!

b) What is the speed of the box after a path of $ s = 5m $?

c) What would be the minimum force application height $ h $ at which the box begins to tip over?

We first draw the free body picture:

Free body picture: box

We have given the weight of the box (directed vertically downwards and attacking the center of gravity) and the normal force $ N $, which reflects the horizontal plane. This prevents the box from being accelerated towards the center of the earth. The normal force is always perpendicular to the plane under consideration. Since we have a horizontal plane here, the normal force is directed vertically (at a 90 ° angle to the plane).

Exercise part a)

We have given forces in $ x $ and in $ y $ directions. In such a case, the equation of motion is broken down into its $ x $ and $ y $ components:

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$ F_x = m a_x $

$ F_y = m a_y $

$ F_x $ is the sum of all forces in the $ x $ direction and $ F_y $ is the sum of all forces in the $ y $ direction that act on the box.

We still have to define a direction for the $ x $ axis and a direction for the $ y $ axis. Basically, the $ x $ axis is always placed in the direction of the movement and the $ y $ axis in the direction of the normal force. The $ x $ -axis is assumed to be directed to the left and the $ y $ -axis is assumed to be directed upwards

The sum of all forces in the $ x $ direction then results in:

$ \ leftarrow \; F_x = F $


The sum of all forces in $ y $ direction:

$ \ uparrow \; F_y = N - G $

We put that in the equations of motion above:

$ F = ma_x $

$ N - G = ma_y $

Since there is no movement in $ y $ direction, $ a_y = 0 $ applies:

$ N - G = 0 $

From this equation it can be seen that:

$ N = G = m \ cdot g = 80 kg \ cdot 9.81 \ frac {m} {s ^ 2} = 784.8 N $.

Weight force and normal force are therefore equal. This simply means that the box has a weight of 784.8 N and exerts this force on the horizontal surface. Since the law of interaction applies, the horizontal surface exerts an equal but opposite force on the box.

Due to $ a_y = 0 $, we do not need this equation to calculate our acceleration (later when the friction is taken into account, the normal force is relevant for the calculation of the acceleration).

So we consider our equation of motion in the $ x $ direction:

$ F = ma_x $

We solve this equation for $ a_x $:

$ a_x = \ frac {F} {m} $

And use the values:

$ a_x = \ frac {240 N} {80 kg} = 3 \ frac {m} {s ^ 2} $

The acceleration is $ a = 3 \ frac {m} {s ^ 2} $.

Exercise part b)

 Next we want to determine the speed of the box according to $ s = 5m $. Since there is constant acceleration here, the speed changes over time. The following relationship applies (see chapter Kinematics):

$ a = \ frac {dv} {dt} $

$ dv = a \; dt $

$ \ int_0 ^ v = a \ int_0 ^ t dt $ (box is moved out of the calm $ v_0 = 0 $. The time measurement starts at $ t_0 = 0 $)

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$ v = a \ cdot t $ (1)

However, we have not given the time here, but the way, so we still have to consider the following relationship:

$ v = \ frac {dx} {dt} $

$ dx = v \; dt $

$ \ int_0 ^ x dx = \ int_0 ^ t v dt $

Inserting $ v = a \ cdot t $:

$ \ int dx = \ int a \ cdot t \; dt $

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$ x = \ frac {1} {2} a \ cdot t ^ 2 $ (2)


We must next eliminate the time $ t $. To do this, we convert equation (1) to $ t $ and insert that into equation (2):

$ t = \ frac {v} {a} $

$ x = \ frac {1} {2} \ cdot a \ cdot (\ frac {v} {a}) ^ 2 $

$ x = \ frac {1} {2} \ cdot a \ cdot \ frac {v ^ 2} {a ^ 2} $

Solve for $ v $:

$ v ^ 2 = \ frac {2 \ cdot x \ cdot a ^ 2} {a} $

$ v ^ 2 = 2 \ times x \ times a $

$ v = \ sqrt {2xa} $

Inserting the known values:

$ v = \ sqrt {2 \ times 5 m \ times 3 \ frac {m} {s ^ 2}} = \ sqrt {30} = 5.48 \ frac {m} {s} $

The speed of the box after $ x = 5m $ is $ v = 5.48 \ frac {m} {s} $.

Exercise part c)

Next we want to determine the tipping height of the box. If the force with the height $ h $ acts on the box from the ground, the box will tip over from a certain height. The formula for this is:

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$ G \ cdot l = F \ cdot h $

$ G $ is the weight of the box, $ l $ is the vertical distance from the weight (center of gravity of the box) to the lower edge of the box (last point of contact between the box and the floor). $ F $ is the force that acts on the box (only the horizontal part of the force) and $ h $ is the height from the force applied to the ground.

In our case, a horizontal force is already acting on the box. If the force were to act on the box at an angle $ \ alpha $, only the horizontal component of this force should be taken into account. We now determine the height h:

$ h = \ frac {G \ cdot l} {F} $

The distance $ l $ from the weight $ G $ to the lower left edge (force acts on the right, so the box tilts on the left).

$ l = 0.75m $

The weight acts on the center of gravity of the box. This is a rectangular box, so the center of gravity is in the middle. The box is 1.5 m wide, so from the center of gravity to the lower left edge the dimensions are $ l = 0.75 m $.

$ h = \ frac {80 kg \ cdot 9.81 \ frac {m} {s ^ 2} \ cdot 0.75m} {240 N} $

$ h = 2.45m $

What does that mean for our 0.5 m high box? This box will not tip over no matter where the force attacks. Because the tipping height from the floor vertically upwards is 2.45 m.