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Vlad1618 [11]
3 years ago
9

A uniform bridge 20.0 m long and weighing 400000 N is supported by two pillars located 3.00 m from each end. If a 19600 N car is

parked 8.00 m from one end of the bridge, how much force does each pillar exert?

Physics
2 answers:
Svetradugi [14.3K]3 years ago
7 0

Answer:

P = 238475 N

Q = 181125

Explanation:

From the diagram attached below,

For the bridge to be at equilibrium,

Sum of upward force = sum of downward force

P+Q = 400000+19600

P+Q = 419600.................. Equation 1

Taking  moment about support A

sum of clockwise moment = sum of anti clockwise moment.

400000(10-3)+19600(8-3) = Q(20-6)

2800000+98000 = 16Q

2898000 = 16Q

Q = 2898000/16

Q = 181125 N.

Substitute the value of Q into equation 1

P+181125 = 419600

P = 419600-181125

P = 238475 N.

Hence, support A exert a force of 238475 N and support B exert a force of 181125 N

goldenfox [79]3 years ago
7 0

Answer:

F1 = 212,600N and F2 = 207,000 N

Explanation:

let; F1 = upward pillar force near the left end of bridge

F2 = upward pillar force near the right end of bridge

WB = bridge weight

WC = car weight

Now, the weight of the bridge and that of the car are acting downwards.

Thus, at translational equilibrium, if we resolve the forces, we'll get;

Σy = 0

Thus, F1 + F2 - WB - WC = 0

We are given the weights of the car and bridge as;

WB = 400000N

WC = 19600N

Thus,we now have ;

F1 + F2 - 400000N - 19600N = 0

F1 + F2 = 400000N + 19600N

F1 + F2 = 419,600N - - - - - (eq1)

let the point where F1 exerts its upward force be the pivot point for system torques.

Now, at rotational equilibrium about F1, we have;

F2•(14) - WC•(5) - WB•(7) = 0

Plugging in values for WC and WB to obtain;

F2•(14) - 19600•(5) - 400000•(7) = 0

14F2 - 98000 - 2800000 = 0

14F2 = 2800000 + 98000

14F2 = 2898000

F2 = 2898000/14

F2 = 207,000 N

Now, let's put this for F2 in eq 1

F1 + 207,000N = 419,600N

Subtract 207,000N from both sides.

F1 = 419,600N - 207,000

F1 = 212,600N

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English Translation

The thermos (also known as "Dewar vase") is an extremely useful device to conserve bodies (essentially liquid) at high temperatures, minimizing energy exchanges with the environment, which is generally colder. A thermos contains water at 60 o C. The thermos + water set has a thermal capacity of C = 80 cal / o C. The system is placed on a table and, after a considerable period of time, its temperature decreases to 55 o C. In this case, it is concluded that the system formed by the thermos and the water inside:

a) lost 400 cal. B) gained 404cal. C) lost 4 850 cal. D) gained 4 850 cal. E) did not exchange heat with the external environment.

Solution

When a system's temperature reduces, it is clear to conclude that the system has lost heat or thermal energy. Since temperature is one of clearest indicators of this, this conclusion is airtight and correct.

But, to know the amount of heat lost to the environment, we now do some thermal energy calculations.

Heat transferrred from or to the water and thermos system = c × ΔT

c = heat capacity of the water and thermos system = 80 cal/°C

ΔT = Change in temperature of the water and thermos system = (final temperature) - (initial temperature)

= 55 - 60 = -5°C

Heat transferred = 80 × -5 = -400 cal.

The minus sign shows that the heat is transferred out of the system, that is, the heat is lost from the system.

Hope this Helps!!!

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