Answer to Question #240858 in Molecular Physics | Thermodynamics for Kelani

Question

The system shown in the figure below is used to lift an object of mass m = 17.5 kg.

 A constant downward force of magnitude F is applied to the loose end of the rope such that the hanging object moves upward at constant speed. Neglecting the masses of the rope and pulleys, find the following.

(a) Find the required value of F.

 __N

(b) Find the tensions T₁, T₂, and T₃. (T₃ indicates the tension in the rope which attaches the pulley to the ceiling.)

T₁ =  __N

T₂ = __ N

T₃ = __ N

(c) Find the work done by the applied force in raising the object a distance of 1.70 m.

__ kJ

Answer 1

(a) Schematic diagram that shows the different forces acting on the given system is shown in the following figure.

Here,

T₃

is the tension force in the string that was connected between the fixed point and the fixed pulley,

T₁

is the tension force in the string that was connecting two pulleys,

T₂

is the tension force in the string that connects moving pulley and the load, and mg is the weight of the load.

Apply law of conservation of force by taking load as reference point is,

T_2=mg

Apply law of conservation of force by taking moving pulley as the reference point is,

$T_1 = \frac{T_2}{2} \\ = \frac{mg}{2}$

Apply law of conservation of force by taking fixed pulley as the reference point is,

$T_3=2T_1 \\ =2\frac{mg}{2} \\ = mg$

The required value of the applied force is,

$F=T_1 \\ = \frac{mg}{2}$

Substitute 17.5 kg for m, and 9.80 m/s² for g.

$F = \frac{17.5 \times 9.80}{2} \\ = 85.75 \;N$

Therefore, the applied force is 85.75 N.

(b) From part (a), the tension force

T1 is 85.75 N.

The tension forces T₂, and T₂ possess same value and is equal to the twice of the T₁. That is,

$T_2 = T_3 = mg$

Substitute 17.5 kg for m, and 9.80 m/s² for g.

$T_2=T_3 \\ = 17.5 \times 9.80 \\ = 171.5 \;N$

Therefore, the tension forces T₂, and T₃ are 171.5 N.

(c) The work W done on the load by a constant force during a linear displacement is given as follows:

$W=Fscos θ$

Here, F is the magnitude of the constant force that was applied on the load, and s is the magnitude of the displacement and θ is the angle between the force vector and the displacement vector.

If the load is raised 1.70 m, the loose end of the rope must be pulled downward a distance s = 3.40 m. The work done by the applied force is,

$W=Fscos θ$

Substitute 85.75 N for F, 3.40 m for s, 0° for θ.

J$W = 85.75 \times 3.40 \times cos θ \\ = 291.55 \;J$

Therefore, the amount of work done by the applied force is 291.55 J.