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zloy xaker [14]
3 years ago
14

In an orthogonal cutting operation, if the friction coefficient between the tool and the chip is decreased while the material an

d the rake angle are kept constant, then the shear angle will increase.
True
False
Engineering
2 answers:
levacccp [35]3 years ago
5 0

Answer:

The Correct Answer is False.              

Explanation:

It is logical that the cutting force increases as the depth of cut increases and rake angle decreases. Deeper cuts remove more material, thus requiring a higher cutting force. As the rake angle, α, decreases, the shear angle, φ , decreases and hence shear energy dissipation and cutting forces increase.

Alexus [3.1K]3 years ago
3 0

Answer:

True

Explanation:

- The shear plane angle is given by the following formula (Merchant's Equation):

                             ∅ = 45 + α/2 - β/2

Where,                  α: Rake Angle

                             β: Friction angle

- It is seen from the relation above that shear plane angle depends on friction angle (β) and rake angle (α).

- The friction angle (β) is related by the friction coefficient (u) with the following relation:

                            u = tan ( β )

                            β = arctan (u)

- The expression becomes:

                            ∅ = 45 + α/2 - arctan (u)/2

- Its given that rake angle (α) remains constant, while friction coefficient (u) between the tool and the chip is decreased.

- For the equation to hold true then shear plane angle (∅) must increase.                          

                             

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Answer:

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There are two types of cellular phones, handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles. P
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Answer:

A) P(W) = 0.5

B) P(MF) = 0.3

C) P(H) = 0.6

Explanation:

We are told that there are two types of cellular phones which are handheld phones (H) that you carry and mobile phones (M) that are mounted in vehicles.

Also, Phone calls can be classified by the traveling speed of the user as fast (F) or slow (W).

Thus, the sample space is combination of types and classification we are given and it is written as;

S = {HF, HW, MF, MW}

A) Now, phones can either be fast(F) or slow(W). Thus, we can write;

P(F) + P(W) = 1

We are given P(F) = 0.5

Thus;

0.5 + P(W) = 1

P(W) = 1 - 0.5

P(W) = 0.5

B) Now, from the problem statement, a phone call can either be made with a handheld(H) or mobile(M). Thus the sample space partition is {H, M} and we can express as;

P(H ∩ F) + P(M ∩ F) = P(F)

We are given P[F] = 0.5 and P[HF] = 0.2.

P(H ∩ F) is same as P[HF]

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Thus;

P(M) = P(MW) + P(MF)

P(M) = 0.1 + 0.3

P(M) = 0.4

Now, since cellular phones can either be handheld(H) or Mobile(M), then we can say;

P(H) + P(M) = 1

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