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almond37 [142]
2 years ago
5

A horizontal pole is attached to the side of a building. There is a pivot P at the wall and a chain is connected from the end of

the pole to a point higher up the wall. There is a tension force F in the chain. What is the moment of the force F about the pivot P?
Physics
1 answer:
PtichkaEL [24]2 years ago
6 0

Answer:

Fscos63

Explanation:

Given that a horizontal pole is attached to the side of a building. There is a pivot P at the wall and a chain is connected from the end of the pole to a point higher up the wall. There is a tension force F in the chain. What is the moment of the force F about the pivot P?

Taking the moment from the pivot point P, that means the moment at point p = 0

Then, if we consider the weight mg of the pole, according to the principle of equilibrium : sum of the upward forces equal to the sum of the downward forces.

Therefore, mg = Fsinø ....... (1)

Also, taking moment at point P

Let the length of the pole = s

The length of the weight of the pole = 1/2 S

Fscosø = mgs/2

The distance s will cancel out

2Fcosø = mg ...... (3)

Substitute mg in equation 1 into equation 3

2fcosø = fsinø

F will cancel out

Tanø = 2

Ø = tan^-1(2)

Ø = 63.4 degree

The moment of force F about pivot point P will be

Moment = force × distance

Moment = Fcos63 × S

Moment = Fscos63

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An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th
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Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

      R = Vo² sin (2θ) / g

      sin 2θ = g R / Vo²

      sin 2θ = 9.8 75/35²

      2θ = sin⁻¹ (0.6)

      θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

       X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

       X = Vox t

       t2 = X2 / Vox = X2 / (Vo cosθ)

        t2 = 37.5 / (35 cos 18.4)

        t2 = 1.13 s

With this time we calculate the height at this point

        Y = Voy t - ½ g t²

        Y = 35 sin 18.4   1.13 - ½ 9.8 1,13²

        Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

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A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because
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Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

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v1= 10m/s

Required

The speed of the sticks

We know that  from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

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