Two equal masses (1 and 2) are connected by the rope across a pulley and placed

Question

Two equal masses (1 and 2) are connected by the rope across a pulley and placed on the larger block A as shown on the picture. Find the minimal acceleration a of the block A along the horizontal so that masses 1 and 2 remain stationary with respect to the block A. Coefficient of kinetic friction between each of the masses and block A is mu, rope is massless and non-stretchable.

Picture here: http://i45.tinypic.com/dzw0sz.jpg

Please explain clearly. I got a = m2g / mu (m1 + mass of block A) but am unsure if it is correct. Thanks.

Explanation / Answer

from force diagram of body 1,
N1=mg
ma=T+uN1

from force diagram of body 2,

N2=ma

mg=T+uN2

Solving we get,

a=g(1+u)/2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.