The drawing shows a frictionless incline and pulley. The two blocks are connecte
ID: 2205451 • Letter: T
Question
The drawing shows a frictionless incline and pulley. The two blocks are connected by a wire (mass per unit length = 0.0195 kg/m) and remain stationary. A transverse wave on the wire has a speed of 86.4 m/s. Neglecting the weight of the wire relative to the tension in the wire, find the masses (a) m1 and (b) m2 of the blocks.
The drawing shows a frictionless incline and pulley. The two blocks are connected by a wire (mass per unit length = 0.0195 kg/m) and remain stationary. A transverse wave on the wire has a speed of 86.4 m/s. Neglecting the weight of the wire relative to the tension in the wire, find the masses (a) m1 and (b) m2 of the blocks.Explanation / Answer
Using the procedures developed in Chapter 4 for usingNewton's second law to analyze the motion of bodies and neglecting the weight of the wirerelative to the tension in the wire lead to the following equations of motion for the twoblocks:
Fx = F m1 g (sin 30.0°) = 0 -------------------(1)
Fy = F m2 g = 0 ---------------------------(2)
where F is the tension in the wire. In Equation (1) we have taken the direction of the +x axisfor block 1 to be parallel to and up the incline. In Equation (2) we have taken the directionof the +y axis to be upward for block 2. This set of equations consists of two equations inthree unknowns, m1, m2, and F. Thus, a third equation is needed in order to solve for any ofthe unknowns.
F = ( m / L ) v2 ---------------------------------(3)
Combining Equation (3) with Equations (1) and (2) leads to
(m / L)v 2 m1g sin 30.0° = 0 ----------------(4)
(m / L)v 2 m2 g = 0 ----------------------(5)
Equations (4) and (5) can be solved directly for the masses m1 and m2.
----------------------------------------------------------
m1 = ( m / L ) v2 / g sin 30
= (0.0195 kg/m)(86.4 m/s)2/ (9.80 m/s 2 ) sin 30.0°
= 29.7 kg
----------------------------------------------------------
substituting values into Equation (5),
m2 = (m / L)v 2 / g
= (0.0195 kg/m)(86.4 m/s)2/ (9.80 m/s 2 )
= 14.8 kg
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