Question

If 50 g is suspended at 15 , and 75 g is suspended at 135 , what mass must
be suspended at what angle to balance these two forces?

Answer 1

Answer on Question #55089, Physics / Mechanics | Kinematics | Dynamics

If $50\mathrm{g}$ is suspended at $15{}^{\circ}$ , and $75\mathrm{g}$ is suspended at $135{}^{\circ}$ , what mass must be suspended at what angle to balance these two forces?

Solution:

Force $F_{1} = 50\mathrm{g}$ at $15{}^{\circ}$ :

$$\vec {F} _ {1} = (5 0 \cos 1 5 {}^ {\circ}, 5 0 \sin 1 5 {}^ {\circ}) = (4 8. 3, 1 2. 9 4)$$

Force $F_{2} = 75\mathrm{g}$ at $135{}^{\circ}$ :

$$\vec {F} _ {1} = (7 5 \cos 1 3 5 {}^ {\circ}, 7 5 \sin 1 3 5 {}^ {\circ}) = (- 5 3. 0 3, 5 3. 0 3)$$

The resultant force is

$$\vec {R} = \overrightarrow {F _ {1}} + \overrightarrow {F _ {2}}$$ $$\vec {R} = (4 8. 3 - 5 3. 0 3, 1 2. 9 4 + 5 3. 0 3) = (- 4. 7 3, 6 5. 9 7)$$

$F_{3}$ is the negative of the resultant $F_{1}$ and $F_{2}$ .

So,

$$\overrightarrow {F _ {3}} = - \vec {R}$$ $$\vec {F} _ {3} = (4. 7 3, - 6 5. 9 7)$$

The magnitude of balance force is

$$F _ {3} = \sqrt {4 . 7 3 ^ {2} + (- 6 5 . 9 7) ^ {2}} = 6 6. 1 4 \approx 6 6 \mathrm {g}$$

To find direction

$$\theta = \tan^ {- 1} \left(\frac {F _ {3 y}}{F _ {3 x}}\right) = \tan^ {- 1} \left(\frac {- 6 5 . 9 7}{4 . 7 3}\right) = - 8 5. 9 {}^ {\circ} = 3 6 0 {}^ {\circ} - 8 5. 9 {}^ {\circ} = 2 7 4. 1 {}^ {\circ}$$

Answer: $66\mathrm{g}$ at $274.1{}^{\circ}$ .

http://www.AssignmentExpert.com/