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nasty-shy [4]
2 years ago
7

Would a measured force of (46.5 0.8 N  ) be in agreement with a theoretically calculated force of (48.4 0.6 N  ) ? Show your w

ork.
Physics
1 answer:
OverLord2011 [107]2 years ago
8 0

Answer:

A measured force of (46.5 0.8 N  ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N  )

Explanation:

From the question we are told that

  Measured force is  F_M  =  [46.5 \pm 0.8 \  N ]

   Calculated force is  F_c =  [48.4 \pm 0.6 \  N ]

Generally the measured force in interval form is

     46.5 - 0.8  < F_M  <  46.5 + 0.8

=>  45.7   < F_M  < 47.3

Generally the calculated  force in interval form is

     48.4 - 0.6  < F_c  <  48.4 + 0.6

=>  47.8   < F_M

Generally looking both interval we see that they do not intersect at any point Hence  

A measured force of (46.5 0.8 N  ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N  )      

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1. Two forces act on a box as follows: F1 = 100 N at 01 = 170° and F2 = 75 N
lilavasa [31]

Answer:

a)  F = 64.30 N,  b) θ = 121.4º

Explanation:

Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components

let's use trigonometry

Force F1

          sin 170 = F_{1y} / F₁

          cos 170 = F₁ₓ / F₁

          F_{1y} = F₁ sin 170

          F₁ₓ = F₁ cos 170

          F_{1y} = 100 sin 170 = 17.36 N

          F₁ₓ = 100 cos 170 = -98.48 N

Force F2

          sin 30 = F_{2y} / F₂

          cos 30 = F₂ₓ / F₂

          F_{2y} = F₂ sin 30

          F₂ₓ = F₂ cos 30

          F_{2y} = 75 sin 30 = 37.5 N

          F₂ₓ = 75 cos 30 = 64.95 N

the resultant force is

X axis

          Fₓ = F₁ₓ + F₂ₓ

          Fₓ = -98.48 +64.95

          Fₓ = -33.53 N

Y axis

         F_y = F_{1y} + F_{2y}

         F_y = 17.36 + 37.5

         F_y = 54.86 N

a) the magnitude of the resultant vector

let's use Pythagoras' theorem

         F = Ra Fx ^ 2 + Fy²

         F = Ra 33.53² + 54.86²

         F = 64.30 N

b) the direction of the resultant

let's use trigonometry

        tan θ’= F_y / Fₓ

        θ'= tan^{-1}  \frac{F_y}{F_x}

        θ'= tan⁻¹ (54.86 / (33.53)

        θ’= 58.6º

this angle is in the second quadrant

The angle measured from the positive side of the x-axis is

        θ = 180 -θ'

        θ = 180- 58.6

        θ = 121.4º

5 0
2 years ago
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