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3 years ago

Determine the magnitude of the component of F directed along the axis of AB. Set F = 520 N .

Physics

2 answers:

Soloha48 [4]3 years ago

7 0

Answer:

The component of F along AB is equal to Fcos45

F = 520N

Component along AB = 520cos45

= 367.7N

This is done by rotating the diamonds such that AB is now taken as the x-axis. Then the force F is resolved along AB.

Explanation:

DIA [1.3K]3 years ago

6 0

Answer:

Force AB = 466.2N

Force AC = 380.7N

Complete Question:

The vertical force F acts downward at A on the two membered frame. Determine the magnitude of the component of F directed along the axis of AB . Set F=520 N

Determine the magnitude of the component of F directed along the axis of AC.

Attached is the image of the question.

Explanation:

Given the force distribution in the image attached:

Let X represent force AB and Y represent force AC

F = 520N

Resolving the force distribution to the vertical and horizontal components.

For vertical component (x axis)

F = Xcos45 + Ycos60 .....1

For horizontal component ( y axis)

Xsin45 = Ysin60 .....2

From equation 2

Y = Xsin45/sin60 ....3

Substituting equation 3 into equation 1

F = Xcos45 + (Xsin45/sin60)cos60

F = X ( cos45 + (sin45/sin60)cos60)

X = F/ ( cos45 + (sin45/sin60)cos60)

Substituting F = 520N and solving for X

X = 520/ ( cos45 + (sin45/sin60)cos60)

X = 466.2N

From equation 3.

Y = Xsin45/sin60

Y = 466.2sin45/sin60

Y = 380.7N

Therefore,

Force AB = 466.2N

Force AC = 380.7N

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A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the

jasenka [17]

Answer:

1.8 cm

Explanation:

$m$ = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

$q$ = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

$\Delta V$ =Potential difference through which the ion is accelerated = 215 V

$v$ = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

$(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}$

$r$ = Radius of the path followed by ion

$B$ = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

$qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm$