Finding Normal Force on Dolly & Floor Without Mass of Dolly

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SUMMARY

The discussion focuses on calculating the normal force exerted by the floor on a dolly and the force the dolly exerts on the floor, given the combined mass of the dolly and carton is 23 kg and they accelerate at 0.41 m/s². The relevant equations include F = ma, with the force exerted by the clerk calculated as 11.807 N. The normal force can be determined using the gravitational force acting on the combined mass, which is 225.4 N (23 kg * 9.8 m/s²). The normal force the dolly exerts on the floor equals the normal force the floor exerts on the dolly, confirming Newton's third law of motion.

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jlk
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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
 
Last edited by a moderator:
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Do you see any reason to not consider the dolly and the carton one entity?
 

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