Find the force necessary to keep the cart moving? - moving crates
To move a large crate on a rough ground, so that with a force at an angle of 21 degrees down below the horizontal. Find the force required to initiate movement of the carriage, because the mass of the box is 32 kg and the coefficient of static friction between the casing and the floor is 0.56.
Tuesday, January 12, 2010
Moving Crates Find The Force Necessary To Keep The Cart Moving?
2 comments:
CFST (21)
FSIN (21)
N = mg - FSIN (21)
u = 56
m = 32 kg
Fnet = CFST (21) - U (mg FSIN (21)) = 0
CFST (21) + uFsin (21) = UMG
F [cos (21) + usin (21)] = UMG
UMG = F / [cos (21) + usin (21)]
F = [56 x 32 kg x 9.8 m / s ^ 2] / (1.134266)
F = 154.8N
EDITED ANSWER
The force must overcome the force of friction to start moving car. The force of friction is μ * where fn is the normal (vertical) force the car into the ground. A component of this force is the weight of the car, m * g, but the force is at an angle below the horizontal, which means that it has vertically against severe. The vertical component F * sinθ. Thus, the total normal force
Fn = m * g - F * sinθ
The force F required to move the car
Cosθ F = μ * m * g * - μ * R * sinθ
Solution for F:
F = μ * m * g / (μ * cosθ + sinθ)
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