A = g m_1 m_2/m_1m_2 Eq. 1 Eq. 2 Eq. 3 Eq. 4 A force of 47.22N is required to st
ID: 1535872 • Letter: A
Question
A = g m_1 m_2/m_1m_2 Eq. 1 Eq. 2 Eq. 3 Eq. 4 A force of 47.22N is required to start a 6.35kg box moving across a horizontal floor. What is the coefficient of static friction between the box and the floor? A force of 47.92 N is required to start a 6.34 kg box moving across a horizontal floor. If the 47.92 N force continues, the box accelerates at 0.55 m/s2. what is the coefficient of kinetic friction? A 16.46kg box is released on a 32 degree incline and accelerates down the incline at o.37m/s2. Find the friction force impeding it's motion. A 14.13kg box is released on a 33 degree incline and accelerates down the incline at 0.24m/s2. what is the coefficient of kinetic friction?Explanation / Answer
when we have a static friction, means that no acceleration. and a) will be solved as the following:
F - f = 0
F = f
F = " µs " * N
F = " µs " * N
F = " µs " * m * g =====> " µs " = (F) / (m*g) ====> " µs " = 48 / (5*9.8) ====>" µs " = 0.98 (right)
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for the kinetic friction, it is possible to have an acceleration:
F - " fk " = ma
F - ma = fk
F - ma = " µk " * N
F - ma = " µk " * m * g
[ F - ma ] / [ m * g ] = " µk " =======> " µk " = [ 48 - 5*0.7 ] / [ 5 * 9.8 ] = 0.91
2...Fb = 16.4kg * 9.8 = 160.72N @ 32deg = Force of box.
Fp = 160.72*sin32 = 85.1686N = Force parallel to the plane.
Fv = 160.72*cos32 = 136.298N = Force perpendicular to the plane.
3...Fb = 14.13kg * 9.8 = 138.474N @ 33deg = Force of box.
Fp = 138.474*sin33 = 75.41N = Force parallel to the plane.
Fv = 138.474*cos33 = 116.134N = Force perpendicular to the plane.
Fp - uFv = ma,
75.41 - 116.13u = 14.13 * 0.24,
75.41 - 116.13u = 3.3912,
-116.137u = 72.0188
u = 0.6201 = coefficient of friction.
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