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3 years ago

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A 40 lb box is suspended from two ropes which each make an angle of 45 degrees with the vertical. What is the tension (in pounds

) on each rope?

1 answer:

givi [52]3 years ago

3 0

Answer:

The tension on each rope is 28.28 lb

Explanation:

Given that,

Weight = 40 lb

Angle = 45°

We need to calculate the tension on each rope

Considering the symmetry of the system, the tension of each rope must be same

So,

$T\sin\theta+T\sin\theta=mg$

Put the value into the formula

$2T\sin\theta=mg$

$T\times\sin45=\dfrac{40}{2}$

$T=\dfrac{40\sqrt{2}}{2}$

$T=28.28\ lb$

Hence, The tension on each rope is 28.28 lb.

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A green truck is moving to the right. A red truck is moving to the left with a speed of 6 m/s. The mass of the red truck is 1,00

Contact [7]

Answer:

2 m/s

Explanation:

$m_1$ = Mass of red truck = 1000 kg

$m_2$ = Mass of green truck= 3000 kg

$u_1$ = Initial Velocity of red truck = 6 m/s

$u_2$ = Initial Velocity of green truck

$v$ = Velocity with which they move together = 0

For elastic collision

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow u_2=-\frac{m_1u_1}{m_2}\\\Rightarrow u_2=-\frac{1000\times 6}{3000}\\\Rightarrow u_2=-2\ m/s$

Velocity of the green truck is 2 m/s

8 0

3 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for <em>x</em> :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

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3 years ago

How many layers does the world have?

Lemur [1.5K]

The world has a total of 7 layers

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3 years ago

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When is a hypothesis formed in the process of conducting an experiment?

tangare [24]

Because you need to have a guess to know what to argue or explain in your experiment

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3 years ago

You stand on a bridge above a river and drop a rock into the water below from a height of 25 m. (Assume no air resistance)

Ilia_Sergeevich [38]

PART a)

here when stone is dropped there is only gravitational force on it

so its acceleration is only due to gravity

so we will have

$a = g = 9.8 m/s^2$

Part b)

Now from kinematics equation we will have

$y = v_i t + \frac{1}{2} at^2$

now we have

y = 25 m

so from above equation

$25 = 0 + \frac{1}{2}(9.8 )t^2$

$t = 2.26 s$

Part c)

If we throw the rock horizontally by speed 20 m/s

then in this case there is no change in the vertical velocity

so it will take same time to reach the water surface as it took initially

So t = 2.26 s

Part D)

Initial speed = 20 m/s

angle of projection = 65 degree

now we have

$v_x = vcos\theta$

$v_x = 20 cos65 = 8.45 m/s$

$v_y = vsin\theta$

$v_y = 20 sin65 = 18.13 m/s$

PART E)

when stone will reach to maximum height then we know that its final speed in y direction becomes zero

so here we can use kinematics in Y direction

$v_f - v_y = at$

$0 - 18.13 = (-9.8) t$

$t = 1.85 s$

so it will take 1.85 s to reach the top

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3 years ago