Considering forces in vertical direction, weight of the crate = normal force on crate by floor
Mg = N
Consider forces in horizontal direction, applied force = force of friction
Fapp= fs
Fapp= 930N
Force of static friction ,fs = ms N , where ms is the coefficient of static friction
F= ms N
F = ms Mg
ms = F/ Mg = 930 / ( 150 x 9.8) = 0.633
A horizontal force of 930 N is needed to overcome the force of static friction between...
A large crate is at rest on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.500. A force F⃗ is applied to the crate in a direction 30.0∘ above the horizontal. The minimum value of F required to get the crate to start sliding is 410 N. Part A What is the mass of the crate? Express your answer with the appropriate units.
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Why is the friction equal to the force?
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2) A 35.0 kg crate is being push across a horizontal floor with
a force F = 150 N that makes an angle θ = 29.0° with the horizontal
as illustrated in the figure below. Find the magnitude of the
acceleration of the crate, given that the coefficient of kinetic
friction between the crate and the floor is 0.20.
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(→F), as shown in the figure below....
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