Question

WHEN A FORCE F ACTS ON A BODY OF MASS M , THE ACCELERATION PRODUCED IN THE BODY IS A . IF THREE EQUAL FORCES F1=F2=F3 ACT ON THE SAME BODY , ANGLE B/W F1 AND F2 IS 90 AND F2AND F3 IS 135. THE ACCELERATION PRODUCED IS- ??
FOR FIGURE , REFER DC PANDEY VOLUME 1 PAGE NO. 138

Answer 1

Answer on Question #40891 – Physics – Mechanics

WHEN A FORCE F ACTS ON A BODY OF MASS M, THE ACCELERATION PRODUCED IN THE BODY IS A. IF THREE EQUAL FORCES F1=F2=F3 ACT ON THE SAME BODY, ANGLE B/W F1 AND F2 IS 90 AND F2 AND F3 IS 135. THE ACCELERATION PRODUCED IS-??

Solution:

$$\left| \vec {F} _ {1} \right| = \left| \vec {F} _ {2} \right| = \left| \vec {F} _ {3} \right| = F$$

Formula for the net force:

$$\vec {\mathrm {F}} _ {\mathrm {n e t}} = \vec {\mathrm {F}} _ {1} + \vec {\mathrm {F}} _ {2} + \vec {\mathrm {F}} _ {3}$$

Resultant force along the X-axis:

$$F _ {n e t X} = F _ {2} - F _ {3} \cdot \cos 4 5 {}^ {\circ} = F - \frac {F}{\sqrt {2}} = F \left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)$$

Resultant force along the Y-axis:

$$F _ {n e t Y} = F _ {1} - F _ {3} \cdot \sin 4 5 {}^ {\circ} = F - \frac {F}{\sqrt {2}} = F \left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)$$

Using the Pythagorean Theorem:

$$F _ {n e t} = \sqrt {\left(F _ {n e t X}\right) ^ {2} + \left(F _ {n e t Y}\right) ^ {2}}$$

(1) and (2) in (3):

$$F _ {n e t} = \sqrt {\left(F \left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\right) ^ {2} + \left(F \left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\right) ^ {2}}$$ $$= \sqrt {2 \left(F \left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\right) ^ {2}} = F \left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right) \cdot \sqrt {2} = F (\sqrt {2} - 1)$$

Newton's second law for the body:

$$F _ {n e t} = M A _ {r e s u l t}$$

(4)in(5): (and using equation $\mathrm{F} = \mathrm{MA}$ )

$$\left\{ \begin{array}{c} \mathrm {F} (\sqrt {2} - 1) = \mathrm {M A} _ {\text {r e s u l t}} \\ \mathrm {F} = \mathrm {M A} \end{array} \right.$$ $$\frac {A _ {\text {r e s u l t}}}{A} = \sqrt {2} - 1$$ $$\mathrm {A} _ {\text {r e s u l t}} = \mathrm {A} (\sqrt {2} - 1)$$

Answer: the acceleration produced is equal to $A\left( {\sqrt{2} - 1}\right)$ .