Answer:
Tension in the chains - In a chain drive, technically, you have a closed-chain (which has no end) going around 2 pulley or gears; looking closely you have 2 parallel chains going in opposite direction. If kept in horizontal direction, the one below the other is the slack side and the other the tight side. The tension on the upper or tight side is more than the slack side. So you need to keep in mind to keep your chain drive tight so that there is no loss or rotation or lags.
Sizes of the pulley/gear - The chain will be warped around a pair of pulley or gear. The sizes of these pulley/gear will also determine the efficiency of the chain drive (consider one big and one small)
Number of pulley/gear - If the number of pulley/gear is more and chain wrapped on it with little complexity will result in decrease in efficiency because of extra tension.
Length of the chain drive - You cannot have much too long chain drive. It will make your slack side more heavy because the end are further away. You have to apply more power and possibilities of lag increases decreasing efficiency. In an ideal situation, this won't happen, but this world isn't ideal.
Friction between chains & pulley/gear - If you have studied gears (involving its teeth), you will come to know that there is friction offered on the two meeting surfaces.
Angle of contact - This would have been explained better with a diagram. Although, if you are familiar with the terms you won't have difficulty understanding. Angle of contact is the angle the chain forms with the pulley/gear at the point of contact with the center of the pulley. The angle of contact should not be too small, or else the things will be slippery.
Explanation:
Answer:
(a) 1294.66 m
(b) 88.44°
Explanation:
d1 = 580 m North
d2 = 530 m North east
d3 = 480 m North west
(a) Write the displacements in vector forms





The resultant displacement is given by



magnitude of the displacement

d = 1294.66 m
(b) Let θ be the angle from + X axis direction in counter clockwise

θ = 88.44°
Answer:
970 kN
Explanation:
The length of the block = 70 mm
The cross section of the block = 50 mm by 10 mm
The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN
The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN
By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force
The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa
The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa
The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa
The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN
Answer:
Stainless steel
Explanation:
I will try to order the solutions from the least correct to the most correct.
Since a temperature greater than 200 ° F is required, that is to say approximately 93 ° c, <em>Polycaprolactone</em> is the least indicated. Its melting point is approximately 60 ° C, so it would not serve the required application.
On the other hand we have<em> Untreated aluminum</em>, which although it has a melting point higher than the required one, without a zinc and magnesium treatment it will easily oxidize in a salty environment, so it cannot be used in this choice either.
We have to compare the two steels.
The<em> Mild Steel </em>has a better corrosion resistance than the previous ones, but in a long-term cycle it will end up full of corrosion and therefore its properties will be highly affected.
Finally, we have <em>stainless steel</em>, which, as the name implies, contains in some of its variations chromium, zinc or magnesium in its alloys, which makes it highly resistant to corrosion.
In addition its melting point is above 1500 ° c.
The best choice is stainless steel.