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# Category Archives: Mechanics

## How wheel + Tyre diameter affects a vehicle’s speed

I have been told a larger wheel/tyre decrease acceleration because its harder to turn the wheel making the car rev slower.

But the same person tells me it will increase the top speed.

My question is if the car has a max rev of 8000rpm but only reaches 7000rpm in its highest gear would a smaller wheel/tyre not make it easier for the car to rev higher thus possible increasing the top speed by using a smaller wheel?

(Garry asked)

**Answer**:

**top speed**other conditions remaining the same.

## DECREASING PRESSURE BY INCREASING AREA

The pressure exerted by a given force can be decreased by increasing the area of contact.

# P=F/A

The equation shows that pressure is inversely proportional to area.

## FOUNDATION OF BUILDINGS

The foundation off building is made more broader than the rest of the building so that the pressure exerted by the weight on earth can be decreased.

## SHOULDER STRAPS OF SCHOOL BAGS

The shoulder straps of school bags are made wider and soft so that the pressure exerted by the heavy bags on shoulders can be reduced. You can feel the paining pressure if you try to replace the broad strap with a narrow one.

FEET OF ELEPHANT AND CAMEL

## MILITARY TANK

The Military tanks have a broad chain so that the pressure exerted by the heavy vehicle on ground is a minimum and it can travel even through marshy places

- Heavy Vehicles have more tyres
- When vehicles get struck in marshy places, wooden planks are used for the vehicle to come up.

## The problem of a rock thrown vertically up

**A rock is thrown vertically upward from the ground with an initial speed 15m/s.**

**a. how high does it go b. how much time is required for the rock to reach its maximum height? c. what is the rock’s height at t=2.00s?**

*(Posted by Merhawi)*

**Answer:**

(a)

u=15m/s

a=-10m/s^{2}

v=0 m/s (at the max height)

S=?

v^{2}-u^{2}=2aS

S=v^{2}-u^{2}/2a

=225/20

**=11.25 m**

(b) From the above case

using v=u+at

t=v-u/a=**1.5s**

(c) Use S = ut + 1/2 at^{2}

put t=2s, u = 15m/s, a=-10m/s^{2}

S=30 – 20 = **10 m**

*(If you use g = 9.8 m/s ^{2 }The answers will be slightly different)*

## A simple numerical on momentum

**Car mass is 1200 kg moving velocity is 108 km/hr. On applying brakes velocity reduced to 36 km/hr. Find change in momentum?**

*(posted by Aditi)*

Answer:

m=1200 kg

u=108 x 5/18=30m/s

v=36×5/18=10m/s

Change in momentum = m(v-u) = 1200 x (30-10)=1200 x 20= __24000 kg m/s__

## Numerical Problem from Friction

“The friction coefficient between a road and the tyre of a vehicle is 4/3.find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding,the vehicle going down at a speed of 10m/s is stopped within 5m” Raju Rabha asked

## Numerical Problem from Motion of Connected systems

Two blocks A and B of mass m1 and m2 respectively are kept in contact on a frictionless table.The experimenter pushes the block A from behind so that the blocks accelerate. If the block A exerts a force F on the block B,what is the force exerted by the experimenter on A.

## Confusion with the definition of unit of time and distance.

Ram Kishore Bajpai asked: (The question is posted as such without rectifying errors.)

“Why is the time defined by distance traveled by light in vacuum 1/299,452,758?

Why does the people use a easier number such as 1/300,000,000? Why didn’t 1 second?”

Answer:

It seems that you are confused a little.

The distance traveled by light in vacuum in 1/299,452,758 of a second is defined as one **metre**. This is because light travels 299,452,758 m in one second in vacuum.

The value 299,452,758 m/s is the speed of light calculated by various experiments and universally accepted standard.

But ordinary situations do not demand this much level of accuracy. If you correct the value 299,452,758 m/s to 1 decimal accuracy, you will get 3.0 x 10^{8} m/s (3,00,000 m/s) and is used in ordinary calculations requiring one or two decimal places accuracy.