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Strike441
7 days ago
5

Suppose the rocket is coming in for a vertical landing at the surface of the earth. The captain adjusts the engine thrust so tha

t rocket slows down at the rate of 2.05 m/s2 . A 6.50-kg instrument is hanging by a vertical wire inside a space ship.Find the force that the wire exerts on the instrument.
Physics
1 answer:
Ostrovityanka [2.7K]7 days ago
5 0

Response:

R= 78.32 N

Clarification:

Input includes

Acceleration,a= 2.05 m/s²

Mass, m = 6.5 kg

The force connected to acceleration

F= mass x Acceleration

F=  ma

F= 6.5 x 2.05 N

F= 13.32 N

The force resulting from weight

F' = m g

F' = 6.5 x 10 N (assuming g= 10 m/s²)

F'= 65 N

Consequently, the total net force is the sum of weight force and acceleration force

R= F + F'

R= 65 + 13.32 N

R= 78.32 N

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Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

F_X = F_{cos \ \theta}

Substituting 42° for θ and 87.0° for F

F_X =87.0 \ N \ *cos \ 42 ^\circ

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Meanwhile, the vertical component is:

F_Y = Fsin \ \theta

Again substituting 42° for θ and 87.0° for F

F_Y =87.0 \ N \ *sin \ 42 ^\circ

F_Y =58.21 \ N

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

f = \mu \ N

Where;

\mu is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting F_Y \ for \ F_{sin \ \theta}. Normal reaction becomes:

N = mg \ - \ F_Y

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

F_X = \mu \ ( mg - \ F_Y)

Rearranging the equation above to isolate \mu leads to:

\mu \ = \ \frac{F_X}{mg - F_Y}

Substituting in the following values:

F_X \ = \ 64.65 \ N

m = 73 kg

g = 9.8 m/s²

F_Y = \ 58.21 N

Thus:

\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}

\mu = 0.0984

Therefore, the coefficient of friction is = 0.0984

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