Asked by nate
To move a large crate across a rough floor, you push on it with a force F at an angle of 21¨¬ below the horizontal. The mass of the crate is 31 kg, ¥ìs is 0.57, and ¥ìk is 0.45. Find the force necessary to start the crate moving
Answers
Answered by
Elena
x: Fcosα –F(fr) =0
y: N-mg-Fsinα = 0 =>
N= mg+Fsinα
Fcosα =F(fr) =μ⒦•N = μ⒦• (mg+Fsinα)
F= μ⒦•m•g/ (cosα - μ⒦•sinα) =
=0.45•31•9.8/(cos40-0.45•sin40) = 286.7 N
(this is the force that necessary to move)
F= μ⒮•m•g/ (cosα - μ⒮•sinα) =
=0.57•31•9.8/(cos40-0.57•sin40) = 433.3 N
(this is the force that necessary to start the motion)
y: N-mg-Fsinα = 0 =>
N= mg+Fsinα
Fcosα =F(fr) =μ⒦•N = μ⒦• (mg+Fsinα)
F= μ⒦•m•g/ (cosα - μ⒦•sinα) =
=0.45•31•9.8/(cos40-0.45•sin40) = 286.7 N
(this is the force that necessary to move)
F= μ⒮•m•g/ (cosα - μ⒮•sinα) =
=0.57•31•9.8/(cos40-0.57•sin40) = 433.3 N
(this is the force that necessary to start the motion)
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