Finding Normal Force on Dolly & Floor Without Mass of Dolly

AI Thread Summary
To find the normal force exerted by the floor on the dolly and the force the dolly exerts on the floor, the combined mass of the dolly and carton (23 kg) and the acceleration (0.41 m/s²) are essential. The normal force can be calculated using the equation F = ma, considering gravitational force as well. The normal force from the floor on the dolly is equal to the weight of the dolly and carton minus any vertical forces acting on it. Ultimately, the normal force exerted by the dolly on the floor is equal in magnitude but opposite in direction to the normal force from the floor on the dolly. Understanding these relationships is crucial for solving the problem accurately.
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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

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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Do you see any reason to not consider the dolly and the carton one entity?
 
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