Physics, 21.10.2020 04:01 Shobhoob

Find the weight of a pumpkin hanging from two scales, if scale one reads 55 N, scale two reads 58 N, and the angle θ between the strings coming from the two scales is 121 degrees.

Find the weight of a pumpkin hanging from two scales, if scale one reads 55 N, scale two reads 58 N

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Answer from: juliopejfuhrf7541

The weight of the pumpkin hanging from the two scales is 98.36 N.

The given parameters;

reading on scale one, a = 55 Nreading on scale two, b = 58 Nangle between the two scales, θ = 121⁰

The weight of the pumpkin hanging from the two scales is equal to the resultant of the two scale readings.

The resultant of the two force is calculated as follows;

R² = a² + b² - 2abcos(θ)

R² = 55² + 58² - 2(55)(58)cos(121)

R² = 9674.7

$R = \sqrt{9674.7} \\\\R = 98.36 \ N$

Thus, the weight of the pumpkin hanging from the two scales is 98.36 N.

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Answer from: andrecoral105

The weight of the pumpkin is 46.793 newtons.

Explanation:

We consider the string-pumpkin system as our system, in which weight of pumpkin create two distinct tensions through cords. The equations of equilibrium are now described:

$\Sigma F_{x} = -T_{1}\cdot \sin \theta_{1}+T_{2}\cdot \sin \theta_{2} = 0$ (Eq. 1)

$\Sigma F_{y} = T_{1}\cdot \cos \theta_{1}+T_{2}\cdot \cos \theta_{2} -W = 0$ (Eq. 2)

Where:

$T_{1}$ , $T_{2}$ - Tension through cords, measured in Newtons.

- Weight of the pumpkin, measured in Newtons.

$\theta_{1}$ , $\theta_{2}$ - Inclination of each cord with respect to the vertical, measured in sexagesimal degrees.

If we know that $T_{1} = 55\,N$ , $T_{2} = 58\,N$ , $\theta_{1} = \theta$ and $\theta_{2} = 121^{\circ}-\theta$ , the system of equations becomes:

$-55\cdot \sin \theta +58\cdot \sin (121^{\circ}-\theta) = 0$ (Eq. 1b)

$55\cdot \cos \theta + 58\cdot \cos (121^{\circ}-\theta) = W$ (Eq. 2b)

From Trigonometry, we remember the following identity:

$\sin (121^{\circ}-\theta) = \sin 121^{\circ}\cdot \cos \theta -\cos 121^{\circ}\cdot \sin \theta$

$\sin (121^{\circ}-\theta) = 0.857\cdot \cos \theta -0.515\cdot \sin \theta$ (Eq. 3)

By applying (Eq. 3) in (Eq. 1b), we get the following expression after expanding and simplifying algebraically:

$-55\cdot \sin \theta + 58\cdot (0.857\cdot \cos \theta - 0.515\cdot \sin \theta) = 0$

$-84.87\cdot \sin \theta +49.706\cdot \cos \theta = 0$

And we solve the equation for $\theta$ :

$\tan \theta = \frac{49.706}{84.87}$

$\theta = \tan^{-1} 0.585$

$\theta \approx 30.328^{\circ}$

Then, the weight of the pumpkin is: ( $\theta \approx 30.328^{\circ}$ )

$W = 55\cdot \cos 30.328^{\circ}+58\cdot \cos (121^{\circ}-30.328^{\circ})$

$W = 46.793\,N$

The weight of the pumpkin is 46.793 newtons.

Answer from: Quest

15.3 cos 65 degrees trig

Answer from: Quest

Is a vector quantity that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.