The figure shows an overhead view of a 0.026 kg lemon hälf änd two of the threforces that act on it as it is ona frictionless table. Force F has a magnitude of 3 N and is ate, -31. Force F2 has a magnitude of 10 N and is at 02- 33". In unit-vector notation, what is thethird force if the lemon half (a) is stationary, (b) has the constant velocity V (137-14) m/s,and (c) has the V = (12ti – 14ij) m/s?, where t is time?%3D

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Description: The figure shows an overhead view of a 0.026 kg lemon hälf änd two of the threforces that act on it as it is ona frictionless table. Force F has a magnitude of 3 N and is ate, -31. Force F2 has a magnitude of 10 N and is at 02- 33". In unit-vector n

(a) The components of force F1 is as follows: F1x=-F1cosθ1F1x=-3 Ncos31°F1x=-2.57 N…

Answered: The figure shows an overhead view of a…

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter6: Applications Of Newton's Laws

Section: Chapter Questions

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The figure shows an overhead view of a 0.026 kg lemon hälf änd two of the thre
forces that act on it as it is ona frictionless table. Force F has a magnitude of 3 N and is at
e, -31. Force F2 has a magnitude of 10 N and is at 02- 33". In unit-vector notation, what is the
third force if the lemon half (a) is stationary, (b) has the constant velocity V (137-14) m/s,
and (c) has the V = (12ti – 14ij) m/s?, where t is time?
%3D

Expert Solution

Step 1

(a) The components of force F₁ is as follows:

$\begin{array}{l} F_{1 x} = - F_{1} \cos θ_{1} \\ F_{1 x} = - (3 N) (\cos 31 °) \\ F_{1 x} = - 2.57 N [1] \\ F_{1 y} = F_{1} \sin θ_{1} \\ F_{1 y} = (3 N) (\sin 31 °) \\ F_{1 y} = 1.54 N [2] \end{array}$

The components of the force F₂ is as follows:

$\begin{array}{l} F_{2 x} = F_{2} \cos θ_{2} \\ F_{2 x} = (10 N) (\sin 33 °) \\ F_{2 x} = 5.45 N [3] \\ F_{2 y} = - F_{2} \sin θ_{2} \\ F_{2 y} = - (10 N) (\cos 33 °) \\ F_{2 y} = - 8.39 N [4] \end{array}$

Step 2

If the lemon half is at stationary, the net force acting on the object is zero. The sum of the force F₁ and F₂ is as follows: $\begin{array}{l} F_{12, x} = F_{1 x} + F_{2 x} \\ F_{12, x} = - 2.57 N + 5.45 N \\ F_{12, x} = 2.88 N [5] \\ F_{12, y} = F_{1 y} + F_{2 y} \\ F_{12, y} = 1.54 N - 8.39 N \\ F_{12, y} = - 6.85 N [6] \end{array}$

The third force is as follows:

${\hat{F}}_{3} = (- 2.88 N) \hat{i} + (6.85 N) \hat{j} [7]$

(b) Since the constant velocity $\vec{v} = (13 \hat{i} - 14 \hat{j}) m / s$ . The force acting on the lemon half is zero. So, the third force is same as in part (a).

${\hat{F}}_{3} = (- 2.88 N) \hat{i} + (6.85 N) \hat{j} [8]$ $\vec{v} = (13 t \hat{i} - 14 t \hat{j}) m / s^{2}$

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