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?

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 xx- and yy-components of the object's acceleration, we need to use Newton's Second Law, which states:

F=ma\vec{F} = m \vec{a} where:

  • F\vec{F} is the net force,
  • mm is the mass of the object (1.2 kg), and
  • a\vec{a} is the acceleration.

We will break the forces into their xx- and yy-components and use them to calculate the acceleration components.

1. Summing the forces in the xx-direction:

From the diagram, the forces acting in the xx-direction are:

  • A force of 4.0N4.0 \, \text{N} to the left (negative xx-direction),
  • A force of 2.0N2.0 \, \text{N} to the right (positive xx-direction).

The net force in the xx-direction is: Fx=4.0N+2.0N=2.0NF_x = -4.0 \, \text{N} + 2.0 \, \text{N} = -2.0 \, \text{N}

2. Summing the forces in the yy-direction:

In the yy-direction, we have:

  • A downward force of 3.0N3.0 \, \text{N},
  • An upward force of 3.0N3.0 \, \text{N},
  • A diagonal force of 3.0N3.0 \, \text{N} at an angle of 45 degrees above the negative xx-axis.

The vertical component of the diagonal force (at 45 degrees) is: Fdiag,y=3.0N×sin(45)=3.0N×222.12NF_{\text{diag}, y} = 3.0 \, \text{N} \times \sin(45^\circ) = 3.0 \, \text{N} \times \frac{\sqrt{2}}{2} \approx 2.12 \, \text{N}

The net force in the yy-direction is: Fy=3.0N+2.12N=0.88NF_y = -3.0 \, \text{N} + 2.12 \, \text{N} = -0.88 \, \text{N}

3. Calculating acceleration:

Now, applying Newton's second law F=ma\vec{F} = m \vec{a} for each direction:

In the xx-direction:

ax=Fxm=2.0N1.2kg=1.67m/s2a_x = \frac{F_x}{m} = \frac{-2.0 \, \text{N}}{1.2 \, \text{kg}} = -1.67 \, \text{m/s}^2

In the yy-direction:

ay=Fym=0.88N1.2kg0.73m/s2a_y = \frac{F_y}{m} = \frac{-0.88 \, \text{N}}{1.2 \, \text{kg}} \approx -0.73 \, \text{m/s}^2

Final Answer:

  • ax=1.67m/s2a_x = -1.67 \, \text{m/s}^2 (to the left),
  • ay=0.73m/s2a_y = -0.73 \, \text{m/s}^2 (downward).

Let me know if you want further details or have any questions.

Related Questions:

  1. How is the resultant acceleration of the object calculated?
  2. What is the magnitude and direction of the net force acting on the object?
  3. How does the angle of the diagonal force affect the components of acceleration?
  4. What would happen to the acceleration if the mass of the object were doubled?
  5. 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 xx- and yy-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