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To move a large crate across a rough floor, you push on it with a force F at an angle of 21° below the horizon?
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, as shown in Figure 6-21. Find the force necessary to start the crate moving, given that the mass of the crate is m = 26 kg and the coefficient of static friction between the crate and the floor is 0.56.
2 Answers
- 5 years agoFavorite Answer
use this equation:
f*cos (x) - ( u * f * sin(x)) = m * g
re-writing that... you get
f = [ (u)(m)(g) ] / [ cos(x) - (u)*sin(x) ]
your answer will be>>> 194.69
- RealProLv 75 years ago
Normal force = weight of crate + vertical component of the force acting down.
= (26)(9.8) + F*sin21
Static friction force = Normal force * 0.56
The horizontal component of the force, Fcos21, has to overcome friction.
Fcos21 = 0.56*(26*9.8 + Fsin21)
Solve for F.
F = (26)(9.8) / (cos21/0.56 - sin21) Newtons
