The forces on mass m1 are the upward tension T as before, plus a downward acceleration m1gsinθ down the inclined plane. In this case, mass m1 will experience an upslo...The forces on mass m1 are the upward tension T as before, plus a downward acceleration m1gsinθ down the inclined plane. In this case, mass m1 will experience an upslope force equal to the tension T and a downslope force m1gsinθ. ΣiFi=m1a⇒T−m1gsinθ−μm1gcosθ=m1aa=m2−m1(μcosθ+sinθ)m1+m2g
Consider the following problem: a block of mass m1 is on a frictionless horizontal surface, and connected by a string, through a pulley, to a mass m2 hanging vertically (Figure \(\PageIn...Consider the following problem: a block of mass m1 is on a frictionless horizontal surface, and connected by a string, through a pulley, to a mass m2 hanging vertically (Figure \PageIndex1). (We assume the string is unbreakable, unstretchable, and of negligible mass.) What is the acceleration of the system?
