Two particles P and Q, each of mass m kg are attached to the ends of a light inextensible string

just in a bit of a pickle with the question any help would be much appreciated

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Two particles P and Q have masses 0.1 kg and 0.5 kg respectively. The particles are attached to the ends of a light inextensible string. Particle P is held at rest on a rough horizontal table. The string lies along the table and passes over a small smooth pulley which is fixed to the edge of the table. Particle Q is at rest on a smooth plane which is inclined to the horizontal at an angle θ, where tan θ = 4/3.The string lies in the vertical plane which contains the pulley and a line of greatest slope of the inclined plane, as shown in Figure 2. Particle P is released from rest with the string taut. During the first 0.5 s of the motion P does not reach the pulley and Q moves 0.75 m down the plane.
(a) Find the tension in the string during the first 0.5 s of the motion.

Daniel D.

asked • 06/12/20

Two particles P and Q have masses m kg and (1 - m) kg respectively. The particles are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. P is held at rest with the string taut and both straight parts of the string vertical. P and Q are each at a height of h meters above horizontal ground. P is released and Q moves downwards. Subsequently Q hits the ground and comes to rest 0.5 seconds later. The max velocity reached by Q is 2m/s.

I've calculated h to be 0.5m. I can't figure out what the masses of the two particles are though. Any help would be greatly appreciated.

1 Expert Answer

Sidney P. answered • 06/12/20

Astronomy, Physics, Chemistry, and Math Tutor

Consider free body diagrams and net forces on the two particles: For P, Fnet = T – mP g = mP a. For Q, Fnet = T – mQ g = - mQ a (negative because its acceleration is downward). Because T has to be the same for both, mP a + mP g = mQ g – mQ a. Solve for a = (mQ – mP) g / (mP + mQ) and substitute their values: a = (1 – 2m) g.

Meanwhile from kinematics vf = vi + a*t, 2 = 0 + a*0.5 and a = 4 m/s2. Δy = -h = vi t + ½ a*t2 = 0 – ½(4)(0.5)2 = -1/2, so height h = 0.5 m.

With a = 4 = (1 – 2m)(9.8), 19.6m = 9.8 – 4 = 5.8 and m = 0.296 kg.

Two particles P and Q, each of mass m kg, are attached to the ends of a light inextensible string. The

string passes over a ﬁxed smooth pulley which is attached to the edge of a rough plane. The plane is

inclined at an angle to the horizontal, where tan =

. Particle P rests on the plane and particle Q

hangs vertically, as shown in the diagram. The string between P and the pulley is parallel to a line of

greatest slope of the plane. The system is in limiting equilibrium.

Show that the coeﬃcient of friction between P and the plane is

A force of magnitude 10 N is applied to P, acting up a line of greatest slope of the plane, and P

accelerates at 2.5 m s

R = mg cos α (R = 9.6m)