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lys-0071 [83]
3 years ago
6

What is the the force available at the roadway surface to perform work?

Engineering
1 answer:
scoray [572]3 years ago
7 0

Answer:

Tractive force or traction

Explanation:

The main purpose of the tractive forces is to improve the ability to <u>transform the engine's energy into the vehicle's movement.</u> There are several systems that have different qualities and uses. Here we explain how they work and what they are for.

In a traction vehicle with one of its axles, its ability to transmit engine power to the ground is limited for two reasons:

-  At least one of the wheels must have adhesion, and this as long as it has a self-locking differential. Otherwise, simply with one lacking grip, we can no longer move forward.

- If there are two wheels that must distribute the power, it will always be easier to saturate the traction capacity of the tire than if we divide the force by four. The example is very simple: if we try to drag an object on the ground by pulling a rubber, it will stretch more than if we pull 4 identical tires, although the force we make is the same.

Contrary to what one might think, all-wheel drive vehicles are nothing recent. What happens is that it did take time to reduce the weight and size, as well as to increase the resistance of the homokinetic joints (they are articulations on the axles to allow the wheel to go up and down with the suspension or turn with the steering) to to adapt these systems to cars.

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Llana [10]

The risks which are associated with such working environment and not reporting accidental breakages of tools is pollution and damage to the body system of individuals.

<h3>What is Risk?</h3>

This is defined as the possibility of something causing a bad event to occur and is common in workplace when tools are left carelessly which can cause injuries etc.

In the case of a working environment with oil spills, it can lead to slipping of the individuals present in the area and not reporting accidental breakages of tools or equipment could also lead to pollution and different forms of accident which thereby making it the most appropriate choice in this context.

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3 0
2 years ago
If the maximum allowable shear stress is 70 MPa, find the shaft diameter needed to transmit 40 kW when the shaft speed is 250 rp
victus00 [196]

Answer:

The diameter is 50mm

Explanation:

The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.

T=(P×60)/(2×pi×N)

T is the Torque

P is the the power to be transmitted by the shaft; 40kW or 40×10³W

pi=3.142

N is the speed of the shaft; 250rpm

T=(40×10³×60)/(2×3.142×250)

T=1527.689Nm

Diameter of a shaft can be obtained from the formula

T=(pi × SS ×d³)/16

Where

SS is the allowable shear stress; 70MPa or 70×10⁶Pa

d is the diameter of the shaft

Making d the subject of the formula

d= cubroot[(T×16)/(pi×SS)]

d=cubroot[(1527.689×16)/(3.142×70×10⁶)]

d=0.04808m or 48.1mm approx 50mm

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4 years ago
Plz answer all of these questions!
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Explanation:

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