Only two forces act on an object (mass=4.19 kg), as in the drawing. Find (a) the magnitude and (b) the direction (relative to the x axis) of the acceleration of the object. +V i I (a) Number i Units 1 145.0⁰ 40.0 N 60.0 N m/s^2 11+x
Only two forces act on an object (mass=4.19 kg), as in the drawing. Find (a) the magnitude and (b) the direction (relative to the x axis) of the acceleration of the object. +V i I (a) Number i Units 1 145.0⁰ 40.0 N 60.0 N m/s^2 11+x
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Problem Statement:
Only two forces act on an object (mass = 4.19 kg), as in the drawing. Find (a) the magnitude and (b) the direction (relative to the x-axis) of the acceleration of the object.
### Diagram Description:
In the provided diagram, we have a rectangular object on a flat surface. Two forces are acting on the object:
- A force of 40.0 N is acting horizontally to the right along the x-axis.
- A force of 60.0 N is acting at an angle of 45.0 degrees to the left of the y-axis (in the second quadrant).
### Solution:
(a) **Magnitude of Acceleration:**
*First, resolve the forces into their x and y components.*
- The horizontal force (40.0 N) acts solely along the x-axis.
- The force at an angle (60.0 N at 45.0 degrees) can be broken down as:
- \( 60.0 \cos(45^\circ) \) in the x-direction.
- \( 60.0 \sin(45^\circ) \) in the y-direction.
*Calculate the components:*
- For force 1 (40.0 N):
- \( F_{1x} = 40.0 \) N
- \( F_{1y} = 0 \) N
- For force 2 (60.0 N):
- \( F_{2x} = 60.0 \cdot \cos(45^\circ) = 60.0 \cdot 0.707 = 42.4 \) N (acting left, so negative)
- \( F_{2y} = 60.0 \cdot \sin(45^\circ) = 60.0 \cdot 0.707 = 42.4 \) N
*Sum of forces in x-direction:*
\[ F_{x\text{total}} = F_{1x} + F_{2x} = 40.0 \text{ N} - 42.4 \text{ N} = -2.4 \text{ N} \]
*Sum of forces in y-direction:*
\[ F_{y\text{total}} = F_{1y} + F_{2y} = 0 \text{ N} + 42.4 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1355fdaa-7203-4437-92a1-5f24f953c1c3%2Fa5f894c3-1e49-414d-9670-15cc151cdd4b%2Fyz8jgla_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
Only two forces act on an object (mass = 4.19 kg), as in the drawing. Find (a) the magnitude and (b) the direction (relative to the x-axis) of the acceleration of the object.
### Diagram Description:
In the provided diagram, we have a rectangular object on a flat surface. Two forces are acting on the object:
- A force of 40.0 N is acting horizontally to the right along the x-axis.
- A force of 60.0 N is acting at an angle of 45.0 degrees to the left of the y-axis (in the second quadrant).
### Solution:
(a) **Magnitude of Acceleration:**
*First, resolve the forces into their x and y components.*
- The horizontal force (40.0 N) acts solely along the x-axis.
- The force at an angle (60.0 N at 45.0 degrees) can be broken down as:
- \( 60.0 \cos(45^\circ) \) in the x-direction.
- \( 60.0 \sin(45^\circ) \) in the y-direction.
*Calculate the components:*
- For force 1 (40.0 N):
- \( F_{1x} = 40.0 \) N
- \( F_{1y} = 0 \) N
- For force 2 (60.0 N):
- \( F_{2x} = 60.0 \cdot \cos(45^\circ) = 60.0 \cdot 0.707 = 42.4 \) N (acting left, so negative)
- \( F_{2y} = 60.0 \cdot \sin(45^\circ) = 60.0 \cdot 0.707 = 42.4 \) N
*Sum of forces in x-direction:*
\[ F_{x\text{total}} = F_{1x} + F_{2x} = 40.0 \text{ N} - 42.4 \text{ N} = -2.4 \text{ N} \]
*Sum of forces in y-direction:*
\[ F_{y\text{total}} = F_{1y} + F_{2y} = 0 \text{ N} + 42.4 \
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