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nordsb [41]
4 years ago
9

14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 65.0°-angle with the horizontal. (a)

Find the horizontal and vertical forces (in N) the ground exerts on the base of the ladder when an 850-N firefighter has climbed 4.10 m along the ladder from the bottom. horizontal
Physics
1 answer:
Mkey [24]4 years ago
4 0

Answer:

The horizontal reaction force is 230.3 N.

Explanation:

Given that,

Length l= 14.0 m

Weight of ladder F = 490 N

Angle = 65°

Weight of firefighter F'= 850 N

Height l'= 4.10 m

Suppose horizontal force magnitude N direction vertical force magnitude N direction

We need to calculate the horizontal reaction force

Horizontal reaction force = normal reaction from wall

Vertical reaction force = weight of ladder +weight of man

F=490+850

F=1340\ N

We need to calculate the moment about bottom is zero

N\times l\sin\theta=F\times\dfrac{l}{2}\cos\theta+F'\times l'\cos\theta

Put the value in the equation

N\times14\sin(65)=490\times7\cos(65)+850\times4.10\cos(65)

N=\dfrac{490\times7\cot(65)+850\times4.10\cot(65)}{14}

N=230.3\ N

Hence, The horizontal reaction force is 230.3 N.

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The question is incomplete. The complete question is :

The Rocket Club is planning to launch a pair of model rockets. To build the rocket, the club needs a rocket body paired with an engine. The table lists the mass of three possible rocket bodies and the force generated by three possible engines.

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Body       Mass (gram)     Engine      Force (newtons)

1                   500                 1                     25

2                  1500                2                    20

3                  750                  3                    30

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Acceleration $=\frac{20}{0.5}$

                      $=\frac{200}{5}$

                      $= 40 \ m/s^2$

Therefore, based on laws of motion of Newton, the Body 1 + Engine 2 combination of the rocket bodies and engines will result in an acceleration of $ 40 \ m/s^2$ at the start of the launch.

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