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If [tex]\( P = 100 \, \text{N} \)[/tex], [tex]\( M = 45 \, \text{kg} \)[/tex], and [tex]\( \theta = 25^\circ \)[/tex], what are the magnitudes for the following forces?

a) The normal force acting on the box [tex]\( = \)[/tex]
b) The frictional force acting on the box [tex]\( = \)[/tex]
c) The weight of the box [tex]\( = \)[/tex]
d) The net horizontal force on the box [tex]\( = \)[/tex]
e) The net vertical force on the box [tex]\( = \)[/tex]



Answer :

GinnyAnswer
Sure, let's solve this step-by-step:

a) The normal force acting on the box:

Given:
- Mass [tex]\( M = 45 \)[/tex] kg
- Angle [tex]\( \theta = 25 \)[/tex] degrees

The normal force is the component of the weight perpendicular to the inclined plane. It is given by:
[tex]\[ \text{Normal force} = M \cdot g \cdot \cos(\theta) \][/tex]

Using the values:
[tex]\[ M = 45 \text{ kg} \][/tex]
[tex]\[ g = 9.81 \text{ m/s}^2 \][/tex]
[tex]\[ \theta = 25 \text{ degrees} \][/tex]

First, convert the angle to radians:
[tex]\[ \theta_{rad} = \frac{25 \times \pi}{180} \][/tex]

Then, calculate:
[tex]\[ \text{Normal force} \approx 400.09 \text{ N} \][/tex]

b) The frictional force acting on the box:

Assuming the coefficient of friction [tex]\( \mu = 0.3 \)[/tex]:

Frictional force is given by:
[tex]\[ \text{Frictional force} = \mu \cdot \text{Normal force} \][/tex]

Using the normal force calculated previously:
[tex]\[ \text{Frictional force} \approx 0.3 \times 400.09 \][/tex]
[tex]\[ \text{Frictional force} \approx 120.03 \text{ N} \][/tex]

c) The weight of the box:

Weight [tex]\( W \)[/tex] is given by:
[tex]\[ W = M \cdot g \][/tex]

Using the values:
[tex]\[ M = 45 \text{ kg} \][/tex]
[tex]\[ g = 9.81 \text{ m/s}^2 \][/tex]

Calculate the weight:
[tex]\[ W \approx 45 \times 9.81 \][/tex]
[tex]\[ W \approx 441.45 \text{ N} \][/tex]

d) The net horizontal force on the box:

Given:
- Applied force [tex]\( P = 100 \)[/tex] N

The net horizontal force is given by:
[tex]\[ \text{Net horizontal force} = P - \text{Frictional force} \][/tex]

Using the frictional force calculated previously:
[tex]\[ \text{Net horizontal force} \approx 100 - 120.03 \][/tex]
[tex]\[ \text{Net horizontal force} \approx -20.03 \text{ N} \][/tex]

The negative sign indicates that the frictional force is greater than the applied force in the opposite direction.

e) The net vertical force on the box:

The net vertical force is given by the component of the weight parallel to the inclined plane:
[tex]\[ \text{Net vertical force} = W \cdot \sin(\theta) \][/tex]

Using the weight calculated previously:
[tex]\[ \text{Net vertical force} \approx 441.45 \times \sin(25^\circ) \][/tex]

Calculate:
[tex]\[ \text{Net vertical force} \approx 186.56 \text{ N} \][/tex]

So, the magnitudes for the given forces are:
a) The normal force [tex]\( \approx 400.09 \text{ N} \)[/tex]
b) The frictional force [tex]\( \approx 120.03 \text{ N} \)[/tex]
c) The weight of the box [tex]\( \approx 441.45 \text{ N} \)[/tex]
d) The net horizontal force [tex]\( \approx -20.03 \text{ N} \)[/tex]
e) The net vertical force [tex]\( \approx 186.56 \text{ N} \)[/tex]

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