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3 years ago

What is the resultant displacement of the two vectors below

Physics

1 answer:

NeX [460]3 years ago

6 0

Answer:

The resultant displacement is 5.8 cm, 31° South of the West axis ⇒ 4th

answer

Explanation:

A resultant vector is the combination of two or more single vectors

If the two vectors are perpendicular, then

The magnitude of the resultant vector = $\sqrt{x^{2}+y^{2}}$ ,where

x represents the horizontal vector and y represents the vertical vector

The direction of the resultant vector is $tan^{-1}\alpha$ , where α

is the angle between the horizontal vector and the resultant vector

Vector 1 is 5 cm due to West-axis

Vector 2 is 3 cm due to South-axis

We need to find the resultant displacement of them

The West and South are perpendicular

The West is the horizontal vector and South is the vertical vector

The magnitude of the resultant displacement s is:

$s=\sqrt{(5)^{2}+(3)^{2}}$

$s=\sqrt{25+9}$

$s=\sqrt{34}=5.8$

The direction of the resultant displacement s is:

$tan^{-1}\frac{3}{5}=31$ ° south of the west (between the west and

the resultant displacement)

The resultant displacement is 5.8 cm, 31° South of the West axis

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4 years ago

Read 2 more answers

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Answer:

$\theta=10.20^{\circ}$

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Explanation:

First of all, we analyze the system of blocks before starting to move.

$\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0$

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$tan(\theta)=0.18$

$\theta=arctan(0.18)$

$\theta=10.20^{\circ}$

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.

$P_{A}sin(\theta)-F_{fA}-F_{spring}=0$

Where:

$F_{spring} = k\Delta l=2.1\Delta l$

$P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0$

$\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}$

$\Delta l=0.10 ft$

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

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