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3 years ago

Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible.

Physics

1 answer:

Kamila [148]3 years ago

8 0

Answer: FR=2.330kN

Explanation:

Write down x and y components.

Fx= FSin30°

Fy= FCos30°

Choose the forces acting up and right as positive.

∑(FR) =∑(Fx )

(FR) x= 5-Fsin30°= 5-0.5F

(FR) y= Fcos30°-4= 0.8660-F

Use Pythagoras theorem

F2R= √F2-11.93F+41

Differentiate both sides

2FRdFR/dF= 2F- 11.93

Set dFR/dF to 0

2F= 11.93

F= 5.964kN

Substitute value back into FR

FR= √F2(F square) - 11.93F + 41

FR=√(5.964)(5.964)-11.93(5.964)+41

FR= 2.330kN

The minimum force is 2.330kN

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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

castortr0y [4]

The vertical component of force exerted by the hi.nge on the beam will be,142.10N.

To find the answer, we need to know more about the tension.

<h3>How to find the vertical component of the force exerted by the hi.nge on the beam?</h3>

Let's draw the free body diagram of the system.
To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.

$F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\$

To find the answer, we have to find the tension,

$Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N$

Thus, the vertical component of the force exerted by the hi.nge on the beam will be,

$F_V=(29*9.8)-(169.43*sin57)=142.10N$

Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.

Learn more about the tension here:

brainly.com/question/28106868

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5 0

1 year ago

Read 2 more answers

spotlight on a boat is 2.5 m above the water, and the light strikes the water at a point that is 8.0 m horizontally displaced fr

ludmilkaskok [199]

Answer:

Explanation:

Let i be the angle of incidence and r be the angle of refraction .

From the figure

Tan ( 90 - i ) = 2.5 / 8

cot i = 2.5 / 8

Tan i = 8 / 2.5 = 3.2

i = 72.65°

From snell's law

sini / sin r = refractive index

sin 72.65 / sinr = 1.333

sin r = .9545 / 1.333

= .72

r = 46⁰

From the figure

Tan r = d / 4

Tan 46 = d /4

d = 4 x Tan 46

= 4 x 1.0355

=4.14 m .

3 0

3 years ago

A 53 kg crate is at rest on a level floor, and the coefficient of kinetic friction is 0.36. The acceleration of gravity is 9.8 m

tino4ka555 [31]

Answer:

42.6 m

Explanation:

mass of crate m = 53 kg

coefficient of kinetic friction, μ = 0.36

acceleration due to gravity, g = 9.8 m/s^2

Force, F = 372.098 N

Net force, f = F - friction force

f = 372.098 - μ m x g = 372.098 - 0.36 x 53 x 9.8

f = 185.114 N

acceleration, a = f / m = 185.114 / 53 = 3.49 m/s^2

initial velocity, u = 0

time, t = 4.94 s

s = ut + 1/2 at^2

s = 0 + 1/2 x 3.49 x 4.94 x 4.94

s = 42.6 m

6 0

3 years ago

The blades in a blender rotate at a rate of 7700 rpm. when the motor is turned off during operation, the blades slow to rest in

Tpy6a [65]

Angular acceleration = (change in angular speed) / (time for the change)

Change in angular speed = (speed at the end) - (speed at the beginning)

For this fan, speed at the end = 7700 rpm, speed at the end = 0 .

Change in angular speed = -7700 rpm

Angular acceleration = (-7700 rpm) / (2.5 sec)

<em>Angular acceleration = -3,080 rev per minute / sec</em>

That's a perfectly good and true answer to the question, but the units are ugly. We really need to fix the units, and convert them into something prettier before we hand in this assignment.

1 rev = 2π radians, and

1 minute = 60 seconds .

Angular acceleration =

(-3,080 rev/min-sec) · (2π rad/rev) · (1 min/60 sec)

AngAccel = (-3,080 · 2π · 1 / 60) · (rev·rad·min / min·sec·rev·sec)

AngAccel = ( -102 and 2/3 · π) · (rad/s²)

<em>AngAccel = -322.5 radian/s²</em>

7 0

2 years ago

Use the galvanometers to determine the amount and direction of the induced current. Which galvanometer is

docker41 [41]

Answer:

Option B

Explanation:

Looking at the 3 galvanometer readings given above, for galvanometer A, the reading is -2 mA.

For galvanometer B, the reading is 4 mA.

While for galvanometer C, the reading is -5 MA

Thus, option B is correct.

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2 years ago