Finding Normal Force on Dolly & Floor Without Mass of Dolly

In summary, in order to find the normal force exerted by the floor on the dolly (part B) and the force exerted by the dolly on the floor (part C), the combined mass of the dolly and the carton should be taken into account. The force exerted by the clerk (part A) can be found using the equation F=ma and taking into account the angle of the force of the man on the carton.
  • #1
jlk
5
0
I'm uncertain how to find the normal force the floor exerts on the dolly (part B) and the force the dolly exerts on the floor (part C) because the mass of the dolly was not given. ONly the combined mass of the dolly and carton was given.

Homework Statement


A stock clerk pushes a carton on a dolly. Refer to image
http://img514.imageshack.us/img514/5184/onewc4.png [Broken]

The carton and dolly have a combined mass of 23 kg. The carton and dolly undergo an acceleration of 0.41 m/s2.
(a) Find the magnitude of the force exerted by the clerk.
(b) Find the magnitude of the normal force that the floor exerts on the dolly.
(c) Find the magnitude of the normal force that the dolly exerts on the floor.


carton and dolly Mass = 23 kg
carton and dolly A = .41 m/s2

g = 9.8m/s2


Homework Equations


F = ma
Fx = max [in x direction]
Fy= may [in y direction]

fmc = force of man on carton
fmcx = fmc* cos 37 [in x direction]
fmcy = fmc* sin 37 [in y direction]


The Attempt at a Solution




Part A) SLOVE for FMC

fmc = force of man on carton

m in this case is the combine mass of the dolly and carton = 23 kg.

Fx = max [in x direction]

fmc = (23kg * 0.41 m/s2) / cos 37
fmc = 11.807 N
 
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  • #2
Do you see any reason to not consider the dolly and the carton one entity?
 
  • #3


So the force exerted by the clerk is 11.807 N.

As a scientist, it is important to always have clear and accurate data in order to make accurate calculations and conclusions. In this case, it is necessary to have the mass of the dolly in order to accurately determine the normal force between the dolly and the floor. Without this information, it is not possible to accurately solve for the normal forces in parts B and C.

One possible solution could be to make assumptions about the mass of the dolly based on its size and material, but this may not be a precise or reliable method. It would be best to obtain the mass of the dolly from the manufacturer or by physically weighing it.

Without the mass of the dolly, it is not possible to accurately determine the normal forces in this scenario. Therefore, it is important to always have complete and accurate data in order to make accurate calculations and conclusions in scientific experiments.
 

1. What is normal force?

Normal force is the force exerted by a surface on an object in contact with it. It is perpendicular to the surface and prevents the object from sinking into the surface.

2. How do you find the normal force on a dolly and floor without knowing the mass of the dolly?

To find the normal force, you can use the equation F=ma, where F is the force, m is the mass, and a is the acceleration. Since the dolly is not accelerating, its acceleration is 0, so the equation becomes F=0. Therefore, the normal force is equal to 0, regardless of the mass of the dolly.

3. Why is it important to know the normal force on the dolly?

The normal force is important because it helps us understand the forces acting on the dolly and how it interacts with the floor. It also allows us to calculate the net force and determine if the dolly is in equilibrium or not.

4. Can the normal force ever be negative?

No, the normal force cannot be negative. It always acts in the direction perpendicular to the surface and opposes the weight of the object in contact with the surface. If it were negative, it would mean that the object is sinking into the surface, which is not possible.

5. How does the normal force affect the weight of the dolly?

The normal force and the weight of the dolly are equal in magnitude but opposite in direction. This means that the normal force reduces the weight of the dolly, making it easier to move or lift.

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