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Stolb23 [73]
4 years ago
5

A block of gelatin is 120mm by 120mm by 40mm whrn unstressed. A force of 49N is applied tangentially to the upper surface causin

g a 10mm displacement relative to the lower surface. The block is placed such that 120mm by 120mm comes on the lower and upper surface, find the shear modulus, shearing stress, shearing strain​
Physics
1 answer:
Inessa05 [86]4 years ago
6 0

Answer:

The shearing stress is 10208.3333 Pa

The shearing strain is 0.25

The shear modulus is 40833.3332 Pa

Explanation:

Given:

Block of gelatin of 120 mm x 120 mm by 40 mm

F = force = 49 N

Displacement = 10 mm

Questions: Find the shear modulus, Sm = ?, shearing stress, Ss = ?, shearing strain​, SS = ?

The shearing stress is defined as the force applied to the block over the projected area, first, it is necessary to calculate the area:

A = 40*120 = 4800 mm² = 0.0048 m²

The shearing stress:

Ss=\frac{F}{A} =\frac{49}{0.0048} =10208.3333Pa

The shearing strain is defined as the tangent of the displacement that the block over its length:

SS=tan\theta =\frac{Displacement}{L}  =\frac{10}{40} =0.25

Finally, the shear modulus is the division of the shearing stress over the shearing strain:

Sm=\frac{10208.3333}{0.25} =40833.3332Pa

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Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

\mu_{sB}=0.126

\mu_{sC}=0.168

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

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\sum \tau_{A}=0

N(3m)-W(1.5m)=0

When solving for N we get:

N=\frac{W(1.5m)}{3m}

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Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

\sum F_{y}=0

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P_{x}=360N(\frac{4}{5})=288N

and finally the torques about C.

\sum \tau_{C}=0

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f_{sB}=\frac{288N(0.75m)}{1.75m}

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With the static friction force in point B we can find the coefficient of static friction in B:

\mu_{sB}=\frac{f_{sB}}{N}

\mu_{sB}=\frac{123.43N}{981N}

\mu_{sB}=0.126

And now we can find the friction force in C.

f_{sC}=P_{x}-f_{xB}

f_{sC}=288N-123.43N=164.57N

f_{sC}=N_{c}\mu_{sC}

and now we can use this to find static friction coefficient in point C.

\mu_{sC}=\frac{f_{sC}}{N}

\mu_{sC}=\frac{164.57N}{981N}

\mu_{sB}=0.168

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