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bija089 [108]
3 years ago
12

In a plate glass factory, sheets of glass move along a conveyor belt at a speed of 15.0 cm/s. An automatic cutting tool descends

at preset intervals to cut the glass to size. Since the assembly belt must keep moving at constant speed, the cutter is set to cut at an angle to compensate for the motion of the glass. If the glass is 72.0 cm wide and the cutter moves across the width at a speed of 24.0 cm/s, at what angle should the cutter be set
Physics
1 answer:
Leni [432]3 years ago
3 0

Answer:

angle at which the cutter should be set = 32°

Explanation:

We are given;

Speed of conveyor belt; v1 = 15 cm/s

Width;w = 72 cm

Speed of the cutter that moves across the width; v2 = 24 cm/s

Let's calculate time;

Since the width and the rate at which the cutter works per seconds is given, thus;

Time = w/v2

Time = 72/24 = 3 seconds

Now, we know that, velocity = distance/time

Thus, distance = velocity x time

So, distance covered;d = 15 x 3 = 45 cm

Now, for us to calculate the angle the cutter should be set, we would treat it as a triangle.

When the cutter cuts it, it covers a distance d which is perpendicular to the width.

Thus, to calculate the angle made with the width, we'll use trigonometric ratio.

Thus, d/w = tan θ where θ is the angle at which the cutter should be set.

So, 45/72 = tan θ

tan θ = 0.625

θ = tan^(-1)0.625

θ = 32°

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Answer:

(a) \hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}

(b) \theta = 85.44^{\circ}

Solution:

As per the question:

Side of the cube, a = 4.4 cm

Coordinates of the diagonally opposite corner, A = <4.4, 4.4, 4.4> cm

Now,

(a) To calculate the unit vector:

\hat{A} = \frac{\vec{A}}{|A|}

\hat{A} = \frac{4.4\hat{i} + 4.4\hat{j} + 4.4\hat{k}}{\sqrt{()4.4}^{2} + (4.4)^{2} + (4.4)^{2}}

\hat{A} = \frac{4.4(\hat{i} + \hat{j} + \hat{k})}{4.4\sqrt{3}}

\hat{A} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}

(b) To calculate the angle between the two vectors say A and A' is given by:

\vec{A}\cdot \vec{A'} = \vec{A}\vec{A'}cos\theta                      

\theta = cos^{- 1}(\frac{\vec{A}\cdot \vec{A'}}{\vec{A}\vec{A'}})        (1)

Now,

The coordinates of the diagonally opposite corner, A' is <0, 0, 1> cm

Thus

\vec{A'} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}

Now,

Using equation (1) :

\theta = cos^{- 1}(\frac{(\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}})\cdot \hat{k}}{|A||A'|})

|A||A'| = (\sqrt{4.4^{2} +4.4^{2} + 4.4^{2}})(\sqrt{0^{2} + 0^{2} + 0^{2}}) = 7.261

Thus

\theta = cos^{- 1}(\frac{\frac{1}{\sqrt{3}}}{7.261})

\theta = cos^{- 1}(0.07946) = 85.44^{\circ}

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