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:
The answer true I’m guessing. It’s a 50/50 chance
Answer:
(a) 104 N
(b) 52 N
Explanation:
Given Data
Angle of inclination of the ramp: 20°
F makes an angle of 30° with the ramp
The component of F parallel to the ramp is Fx = 90 N.
The component of F perpendicular to the ramp is Fy.
(a)
Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.
Resolve F into its x-component from Pythagorean theorem:
Fx=Fcos30°
Solve for F:
F= Fx/cos30°
Substitute for Fx from given data:
Fx=90 N/cos30°
=104 N
(b) Resolve r into its y-component from Pythagorean theorem:
Fy = Fsin 30°
Substitute for F from part (a):
Fy = (104 N) (sin 30°)
= 52 N
Answer
given,
ω₁ = 0 rev/s
ω₂ = 6 rev/s
t = 11 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
11 α = 6 - 0
= 0.545 rev/s²
The angular displacement
θ₁= ωi t + (1/2) α t²
θ₁= 0 + (1/2) (0.545)(11)^2
θ₁= 33 rev
case 2
ω₁ = 6 rev/s
ω₂ = 0 rev/s
t = 14 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
14 α = 0 - 6
= - 0.428 rev/s²
The angular displacement
θ₂= ωi t + (1/2) α t²
θ₂= 6 x 14 + (1/2) (-0.428)(14)^2
θ₂= 42 rev
total revolution in 25 s is equal to
θ = θ₁ + θ₂
θ = 33 + 42
θ = 75 rev
Explanation
(m) is measured in kilograms (kg)
<h2>(F) is measured in newtons (N)</h2>
<h3>acceleration (a) is measured in metres per second squared (m/s²)</h3>
