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## Homework Statement

A(n) 895 N crate is being pushed across a level floor by a force of 397 N at an angle of 21◦ above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.24. The acceleration of gravity is 9.81 m/s.

What is the acceleration of the box? Answer in units of m/s/s

## The Attempt at a Solution

to find mass, Fg = mg

895 = m9.8

m = 91.3265

I've tried

ΣF

_{y}= F

_{N}- mg - F

_{A}sinθ

ΣF

_{y}= F

_{N}- mg - F

_{A}sinθ = ma

ΣF

_{y}= F

_{N}- mg - F

_{A}sinθ = 0

F

_{N}= mg + 397sin21

F

_{N}= 895 + 142.272

F

_{N}= 1037.27

∑F

_{x}= F

_{A}cosθ - F

_{f}

∑F

_{x}= 397cos21 - (0.24)(1037.27)

397cos21 - (0.24)(1037.27) = ma

397cos21 - (0.24)(1037.27) = (91.3265)a

= 1.33 m/s/s but this is not correct

can anyone tell me where i went wrong/what is the correct way? thanks