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saul85 [17]
3 years ago
6

A uniform shelf 1.8 meters long with mass 15 kg is mounted by a small hinge on a wall. The beam is held in a horizontal position

by a cable that makes an angle of 37°. There is a 4.9-kg round light hanging down and is 0.55 meters from the wall. What is the tension in the supporting cable?
Physics
1 answer:
jeka57 [31]3 years ago
8 0

Answer:

 T = 153.72 N

Explanation:

For this exercise we must use the conditions of translational and rotational equilibrium.

Let's set a frame of reference on the hinge, start by writing the rotational equilibrium relationship, suppose counterclockwise rotation is positive

We look for the components of the cable tension with trigonometry

           cos 37 = Tₓ / T

           sin 37 = T_{y} / T

           Tₓ = T cos 37

           T_{y} = T sin 37

the expression for rotational equilibrium is

        T_{y} L + Tₓ 0 - W L / 2 - W_light  0.55 = 0

where L is the length L=  1.8 m,

        T_{y} = (W L/2 + W_lght 0.55) / L

        T sin 37 = Mg /2 + m_light g 0.55 / L

        T = (M / 2 + m_light 0.55 / L) g / sin 35

 let's calculate

        T = (15/2 + 4.9 0.55 / 1.8) 9.8 / sin 35

         T = 153.72 N

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A car travelling at 15 m/s comes to rest in a distance of 14 m when the brakes are applied.
stira [4]

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3 0
3 years ago
A fan cart with the fan set to high rolled across the floor. The cart's speeds with the fan on high are shown below. If the fan
Vadim26 [7]

Answer:

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Explanation:

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3 0
3 years ago
I need an answer for this plzz!!<br>number 2 <br>anybody can help ??
Anuta_ua [19.1K]
2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a \:  =  \:  \frac{dv}{dt}  \:  =  \:  \frac{25 \:  \frac{m}{s} }{50 \: s} \:  =  \: 0.5 \:  \frac{m}{ {s}^{2} }
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2ax \:  =  \:  {v}^{2}  \:  -  \:  {u}^{2}
x \:  =  \:  \frac{ {v}^{2}  \:   -  \:  {u}^{2} }{2a}  \:  =  \:   \frac{{(25 \:  \frac{m}{s})}^{2}  \:  -  \:  {(0 \:  \frac{m}{s} )}^{2} }{2 \:  \times  \: 0.5 \:  \frac{m}{ {s}^{2} } } \:  =  \: 625 \: m
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
8 0
3 years ago
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