PLEASE HELP ME WITH THIS ONE QUESTION
1 answer:
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Answer:
a = 6.4 [m/s²]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration. In this way, we have the following equation.
∑F = m*a
where:
∑F = sum of forces (horizontal force = 40 [N] and friction force = 8 [N])
m = mass of the block = 5 [kg]
a = acceleration [m/s²]
Now replacing:
Note: the sign of the friction force is negative since this force is acting against the movement of the block.
Answer:
principal stresses :б1 = 32.62mPa б2 = 31.38mPa
Max Shear stress : 16.31 mPa
Orientation of max principle plane = 44.43°
Orientation of minimum principal plane = 134.43°
Explanation:
Given data:
Torque = 50 N-m
weight = 80 kgs
half of weight is subjected to each leg
radius of bone = 10 mm = 0.010 m
<u>a) Determine the principal stresses and shear stress</u>
first calculate the max shear stress ( this will occur in the outermost element
= 16T / π*d^3 where : T = 50 , d = 0.020 m
hence max shear stress = 32 mPa
next determine compressive stress
= ( 40*g) / π/4*d^2 . where : d = 0.020 m , g = 9.81
hence compressive stress = 1.24 mPa
draw and calculate the radius of Mohr's circle
radius of Mohr's circle = 32.0060
Hence principal stresses = 32.0060 ± 0.62
б1 = 32.62mPa
б2 = 31.38mPa
attached below is the remaining part of the solution
Answer:
a)
b)
Explanation:
Given:
height of water in one arm of the u-tube,
a)
Gauge pressure at the water-mercury interface,:
we've the density of the water
b)
Now the same pressure is balanced by the mercury column in the other arm of the tube:
<u>Now the difference in the column is :</u>