Question

Four masses are attached to a shaft at planes A,B,C and D at equal radii. The distance of the planes B,C and D from A are 40 cm, 50 cm and 120 cm respectively. The masses at A,B and C are 60 kg, 45 kg and 70 kg respectively. If the system is in complete balance, determine the mass at Dand position of masses B,C and D with respect to A.

Answer 1

Question #81298

Four masses are attached to a shaft at planes A, B, C and D at equal radii. The distance of the planes B, C and D from A are 40 cm, 50 cm and 120 cm respectively. The masses at A, B and C are 60 kg, 45 kg and 70 kg respectively. If the system is in complete balance, determine the mass at D and position of masses B, C and D with respect to A.

Answer:

For dynamic balance the sum of forces and moments about any plane should be zero. Let us consider the forces and moments about plane D. So, we have the following system of equations (assume the position of mass A is 0 degrees):

$$\sum F _ {x} = 0: 60 + 45 \cos \theta_ {B} + 70 \cos \theta_ {C} + m _ {x} = 0,$$ $$\sum F _ {y} = 0: 45 \sin \theta_ {B} + 70 \sin \theta_ {C} + m _ {y} = 0,$$ $$\sum M _ {D x} = 0: 60 AD + 45 BD \cos \theta_ {B} + 70 CD \cos \theta_ {C} = 0,$$ $$\sum M _ {D x} = 0: 45 BD \sin \theta_ {B} + 70 CD \sin \theta_ {C} = 0,$$

where $\theta_{\mathrm{B}}$ and $\theta_{\mathrm{C}}$ are the positions of the masses B and C, respectively,

$m_{x}$ and $m_{y}$ are the components of the mass D.

The distances are as follow:

$$AD = 120 \mathrm{~cm},$$ $$BD = AD - AB = 120 - 40 = 80 \mathrm{~cm},$$ $$CD = AD - AC = 120 - 50 = 70 \mathrm{~cm}.$$

Substituting into (3) and (4), we can define the following relations:

$$\cos \theta_ {C} = - \frac {7200 + 3600 \cos \theta_ {B}}{4900} = - \frac {72 + 36 \cos \theta_ {B}}{49},$$ $$\sin \theta_ {C} = - \frac {3600 \sin \theta_ {B}}{4900} = - \frac {36 \sin \theta_ {B}}{49}.$$

Let us use the Pythagorean identity:

$$\cos^ {2} \theta_ {C} + \sin^ {2} \theta_ {C} = 1,$$ $$\left(\frac {72 + 36 \cos \theta_ {B}}{49}\right) ^ {2} + \left(\frac {36 \sin \theta_ {B}}{49}\right) ^ {2} = 1,$$ $$72 ^ {2} + 2 \cdot 72 \cdot 36 \cos \theta_ {B} + 36 ^ {2} \left(\cos^ {2} \theta_ {B} + \sin^ {2} \theta_ {B}\right) = 49 ^ {2},$$ $$\cos \theta_ {B} = \frac {49 ^ {2} - 72 ^ {2} - 36 ^ {2}}{2 \cdot 72 \cdot 36} = - 0.7868.$$

Substitute into (6):

$$\cos \theta_ {C} = - \frac {72 + 36 (- 0.7868)}{49} = - 0.8913.$$

Taking into account (4), we see, that $\theta_{B}$ and $\theta_{C}$ have opposite signs. Thus, the positions of the masses B and C are

$$\theta_ {B} = 1 4 1. 9 ^ {\circ}, \quad \text {or} \quad \theta_ {B} = - 1 4 1. 9 ^ {\circ},$$ $$\theta_ {C} = - 1 5 3. 0 ^ {\circ} \quad \text {or} \quad \theta_ {C} = 1 5 3. 0 ^ {\circ}.$$

Substitute into (1) and (2):

$$6 0 + 4 5 (- 0. 7 8 6 8) + 7 0 (- 0. 8 9 1 3) + m _ {x} = 0,$$ $$m _ {x} = 4 5 \cdot 0. 7 8 6 8 + 7 0 \cdot 0. 8 9 1 3 - 6 0 = 3 7. 8 0,$$ $$4 5 \sin 1 4 1. 9 ^ {\circ} + 7 0 \sin (- 1 5 3 ^ {\circ}) + m _ {y} = 0 \quad \text {or} \quad 4 5 \sin (- 1 4 1. 9 ^ {\circ}) + 7 0 \sin 1 5 3 ^ {\circ} + m _ {y} = 0,$$ $$m _ {y} = 3. 9 6 7 \quad \text {or} \quad m _ {y} = - 3. 9 6 7.$$

Thus the mass $D$ and its position are:

$$m = \sqrt {m _ {x} ^ {2} + m _ {y} ^ {2}} = \sqrt {3 7 . 8 ^ {2} + 3 . 9 6 7 ^ {2}} = 3 8 \mathrm {k g},$$ $$\tan \theta_ {D} = \frac {m _ {y}}{m _ {x}},$$ $$\theta_ {D} = \mathrm {a t a n} \frac {3 . 9 6 7}{3 7 . 8} = 6 ^ {\circ} \quad \mathrm {o r} \quad \theta_ {D} = \mathrm {a t a n} \frac {- 3 . 9 6 7}{3 7 . 8} = - 6 ^ {\circ}.$$