Asked by abcdefg
Two crates, of mass 75 kg and 110 kg, are in contact and at rest on a horizontal surface. A 620-N force is exerted on the 75-kg crate. If the coefficient of kinetic friction is 0.15, calculate (a) the acceleration of the system, and (b) the force that each crate exerts on the other.
(c) Repeat with the crates reversed.
(c) Repeat with the crates reversed.
Answers
Answered by
henry2,
a. M = M1+ M2 = 75 + 110 = 185 kg.
M*g = 185 * 9.8 = 1813 N. = Wt. of crates. = Normal force.
0.15*1813 = 272 N. = Force of kinetic friction.
620-272 = M*a.
348 = 185*a,
a =
b. = Mg*sin A = 1813*sin 0 = 0 = Force parallel to the hor. plane. = Force
that each crate exert on the other.
c. Nothing changed.
M*g = 185 * 9.8 = 1813 N. = Wt. of crates. = Normal force.
0.15*1813 = 272 N. = Force of kinetic friction.
620-272 = M*a.
348 = 185*a,
a =
b. = Mg*sin A = 1813*sin 0 = 0 = Force parallel to the hor. plane. = Force
that each crate exert on the other.
c. Nothing changed.
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