For mass 1, Set up your x coordinate so that Tension is in the positive, and so that the Force of Friction and the x component of weight is in the negative. No other forces exist in the x.
Set up your y coordinate perpendicular to the surface of the ramp. Make the normal force positive y, and the y component of weight negative. No other forces exist in the y.
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For mass 2, set up your secondary coordinate system so that straight down is positive x, and straight up is negative x. Thus, the tension is negative, and the weight of the mass is in the positive.
Given this, we have:
Fx1: T-(m1)gsin(theta)-(μk)Fn=(m1)a--->
a= (T-(m1)gsin(theta)-(μk)Fn)/(m1)
Fy1: Fn-(m1)gcos(theta)=0---> Fn=(m1)gcos(theta)
Fx2: (m2)g-T=(m2)a---> T=(m2)(g-a)
Therefore, >>>a=((m2)g-(m1)gsin(theta)-(μk)Fn)/((m1)+(m2))<<<
and, >>>T=(m2)(g-a)<<<
Where, (m1)=Mass 1, (m2)=Mass 2, Fn=Normal Force, and (><) indicates final solution.
In the figure below, m1 = 3.8 kg, m2 = 5.2 kg, and the coefficient of kinetic friction between the inclined plane and the 3.8-kg block is μk = 0.3. Find the magnitude of the acceleration of the masses and the tension in the cord.
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1 answer