Respuesta :
Answer:
The normal force is the force that the floor does as a reaction of the gravitational force that an object does against the floor (is the resistance that objects have when other objects want to move trhough them, and the force comes by the 3rd Newton's law, and this is specially used in cases where the first object is fixed, like walls or the floor). With this in mind, the point in where the normal force will be greater is the point that is closer to the center of mass of the object (the point with more mass)
If the wheels are in the extremes of the object, and the center of mass is in the middle of the object, the normal force will be equal. Now if for example, you put a little mass in one end of the object, now the center of weight displaces a little bit and is not centered, and the side is where you put the weight on will receive a bigger normal force from the floor than the other side.
The force on the front and rear wheels can be determined by finding the
force acting under equilibrium conditions.
The normal reaction between the floor and the front wheel is larger. The
correct option is c) The force on the front wheels must be greater than the
force on the rear wheels.
Reasons:
The forces acting on the wheels are;
Vertical forces acting on the cart:
The weight of the cart acting on the floor
The normal reaction of the floor on the wheels
The horizontal acting on the cart:
Pushing force acting horizontally
Friction force acting in the reverse direction
At equilibrium, the cart does not tip over
Sum of moment about the rear wheel = [tex]\mathbf{W \times \dfrac{d}{2} + F \times h - N_{front} \times d} = 0[/tex]
Where;
h = The height of the cart
d = depth of the cart
[tex]N_{front}[/tex] = The normal reaction at the front wheels
Therefore;
[tex]\displaystyle N_{front} = \frac{W \times \dfrac{d}{2} + F \times h}{d} = \mathbf{ \frac{W}{2} + \frac{F \times h}{d}}[/tex]
Sum of moment about the front wheel = [tex]\mathbf{ F \times h + N_{rear} \times d - W \times \dfrac{d}{2} } = 0[/tex]
Therefore;
[tex]\displaystyle N_{rear} = \frac{W \times \dfrac{d}{2} - F \times h}{d} = \mathbf{ \frac{W}{2} - \frac{F \times h}{d}}[/tex]
Which gives;
[tex]\displaystyle N_{front} = \mathbf{ \displaystyle N_{rear} + 2 \times \frac{F \times h}{d}}[/tex]
[tex]\displaystyle 2 \times \frac{F \times h}{d} > 0[/tex]
Therefore;
[tex]\displaystyle N_{front} > \displaystyle N_{rear}[/tex]
The correct option is c) The force on the front wheels must be greater than
the force on the rear wheels.
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