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2 years ago

7

A frame hanging on a wall is held by two cables. The tension in each cable is 30 N, and the cables make an angle of 45° with the

horizontal, as shown in the picture. What is the weight of the frame?
A) 10.2 N
B) 21.2 N
C) 32.4 N
D) 42.4 N

2 answers:

Vesna [10]2 years ago

6 0

Answer:

(D) 42.4N

Explanation:

Since the frame is at rest, the net force acting on it must be 0. There are three forces acting on it: the gravity and the opposing forces of the two cables.

Since the gravity is a vertical force, we are only interested in the vertical components of the remaining forces. The net force equation is

F_net = 0 = F_g -2 * F_y

The vertical force of one cable (using the information in the drawing) is:

F_y = 30N * sin 45 deg = 21.21N

Now the weight can be determined:

0 = F_g - 2 * F_y

F_g= 2 * F_y = 2 * 21.21N = 42.4N

The weight of the frame is about 42.4N.

DanielleElmas [232]2 years ago

4 0

Vector:T₁+T₂+F=0 , you put the vectors where you need, on the site I'm not going to put them

We design the relationship on the vertical axis Oy:

T₁sinα+T₂sinα=0

T₁=T₂=>2Tsinα=0

We design the relationship on the horizontal axis Ox :

2Tcosα-F=0=>F=2Tcosα=2*30*√2/2=30√2=42.4 N

The right answer is D

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Read 2 more answers

a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde

Svet_ta [14]

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θ/360·2·π·s = 2·π·r

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h = s² - r²

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3·r²·s² - 4·r⁴ = 0

3·r²·s² = 4·r⁴

3·s² = 4·r²

$\underline{\left \right. The \ radius, \, r =\sqrt{ \dfrac{3}{4}} \cdot s}$

$\underline{The \ height, \, h =\sqrt{s^2 - \dfrac{3}{4}\cdot s^2} = \dfrac{s}{2}}}$

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brainly.com/question/14466080

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