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Pepsi [2]
3 years ago
11

Three collinear forces,F1=45N west,F2=63N east and an unknown force F3 are applied to an object.The resultant force of the three

forces is 12N west.
Show by means of a tail to head method of vector addition how the magnitude of F3 could be found?​
Physics
1 answer:
Greeley [361]3 years ago
6 0

Take east to be the positive direction. Then the resultant force from adding <em>F</em>₁ and <em>F</em>₂ is

<em>F</em>₁ + <em>F</em>₂ = (-45 N) + 63 N = 18 N

which is positive, so it's directed east.

To this we add a third force <em>F</em>₃ such that the resultant is 12 N pointing west, making it negative, so that

18 N + <em>F</em>₃ = -12 N

<em>F</em>₃ = -30 N

So <em>F</em>₃ has a magnitude of 30 N and points west.

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Approximately 2.0 \times 10^{-23}\; \rm m \cdot s^{-1} if that athlete jumped up at 1.8\; \rm m \cdot s^{-1}. (Assuming that g = 9.81\; \rm m\cdot s^{-1}.)

Explanation:

The momentum p of an object is the product of its mass m and its velocity v. That is: p = m \cdot v.

Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: v(\text{athlete, before}) = 0 and v(\text{earth, before}) = 0. Therefore:

\begin{aligned}& p(\text{athlete, before}) = 0\end{aligned} and p(\text{earth, before}) = 0.

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\begin{aligned}& p(\text{athlete, after}) + p(\text{earth, after}) \\ & =p(\text{athlete, before}) + p(\text{earth, before}) \\ &= 0\end{aligned}.

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\begin{aligned}& m(\text{athlete}) \cdot v(\text{athelete, after}) \\ &= - m(\text{earth}) \cdot v(\text{earth, after})\end{aligned}.

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\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}.

The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is g = 9.81\; \rm N \cdot kg^{-1}.

\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}.

Calculate v(\text{earth, after}) using m(\text{earth}) and v(\text{athlete, after}) values from the question:

\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \\ &\approx -2.0 \times 10^{-23}\; \rm m \cdot s^{-1}\end{aligned}.

The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately 2.0 \times 10^{-23}\; \rm m \cdot s^{-1}.

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