Math Problem Statement
The forces in the figure act on a 1.2 kg object. What is the value of the x -component and y - component of the object's acceleration?
Solution
To determine the - and -components of the object's acceleration, we need to use Newton's Second Law, which states:
where:
- is the net force,
- is the mass of the object (1.2 kg), and
- is the acceleration.
We will break the forces into their - and -components and use them to calculate the acceleration components.
1. Summing the forces in the -direction:
From the diagram, the forces acting in the -direction are:
- A force of to the left (negative -direction),
- A force of to the right (positive -direction).
The net force in the -direction is:
2. Summing the forces in the -direction:
In the -direction, we have:
- A downward force of ,
- An upward force of ,
- A diagonal force of at an angle of 45 degrees above the negative -axis.
The vertical component of the diagonal force (at 45 degrees) is:
The net force in the -direction is:
3. Calculating acceleration:
Now, applying Newton's second law for each direction:
In the -direction:
In the -direction:
Final Answer:
- (to the left),
- (downward).
Let me know if you want further details or have any questions.
Related Questions:
- How is the resultant acceleration of the object calculated?
- What is the magnitude and direction of the net force acting on the object?
- How does the angle of the diagonal force affect the components of acceleration?
- What would happen to the acceleration if the mass of the object were doubled?
- How do we resolve forces acting at an angle into their components?
Tip:
When dealing with forces at an angle, always break them into their - and -components using trigonometric functions like sine and cosine.
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Math Problem Analysis
Mathematical Concepts
Newton's Second Law
Vector Decomposition
Trigonometry
Formulas
\( \vec{F} = m \vec{a} \)
\( F_{\text{diag}, y} = F \times \sin(\theta) \)
\( a_x = \frac{F_x}{m} \)
\( a_y = \frac{F_y}{m} \)
Theorems
Newton's Second Law
Suitable Grade Level
High School
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