it takes a maximum force of 460 N to move a 125 kg crate. What is the coefficient of the maximum static friction between the crate and the floor?
Wc = mg = 125kg * 9.8N/kg = 1225N =
Weight of crate.
Fc = 1225N @ 0deg.
Fp = 1225sin(0) = 0 = Force parallel to the floor = Hor comp.
Fv = 1225cos(0) = 1225N. = Force perpendicular to the floor.
Fn = Fap - Fp - Fs = 0,
460 - 0 - u*Fv = 0,
u*Fv = 460,
u = 460 / Fv = 460 / 1225 = 0.376 =
coefficient of static friction.
To find the coefficient of maximum static friction between the crate and the floor, we can use the formula:
\(F_{\text{{friction}}} = \text{{coefficient of static friction}} \times F_{\text{{normal}}}\)
Here,
\(F_{\text{{friction}}} = 460 \, \text{{N}}\) (maximum force applied)
\(F_{\text{{normal}}} = \text{{weight}} = m \times g\)
\(m = 125 \, \text{{kg}}\) (mass of the crate)
\(g = 9.8 \, \text{{m/s}}^2\) (acceleration due to gravity)
First, let's calculate the weight of the crate:
\(F_{\text{{normal}}} = m \times g = 125 \, \text{{kg}} \times 9.8 \, \text{{m/s}}^2 = 1225 \, \text{{N}}\)
Now, rearranging the formula and substituting the known values:
\(\text{{coefficient of static friction}} = \frac{F_{\text{{friction}}}}{F_{\text{{normal}}} }\)
\(\text{{coefficient of static friction}} = \frac{460 \, \text{{N}}}{1225 \, \text{{N}}} \approx 0.375 \)
Therefore, the coefficient of maximum static friction between the crate and the floor is approximately 0.375.