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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.
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
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]
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
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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