Answer :
To determine the magnitude and direction of the acceleration induced by the two applied forces, we need to perform a series of steps to find the net force acting on the object and then use Newton's second law to find the acceleration.
Let's break down the process step by step:
1. Determine the components of each force:
\( F_1 = 68 \, \text{N} \) at \( 24^\circ \) North of East.
[tex]\[ F_{1_x} = F_1 \cos(24^\circ) = 62.12109111969686 \, \text{N} \][/tex]
[tex]\[ F_{1_y} = F_1 \sin(24^\circ) = 27.658091729154414 \, \text{N} \][/tex]
\( F_2 = 32 \, \text{N} \) at \( 48^\circ \) North of West.
[tex]\[ F_{2_x} = F_2 \cos(48^\circ) = 21.412179403483464 \, \text{N} \][/tex]
[tex]\[ F_{2_y} = F_2 \sin(48^\circ) = 23.780634415276616 \, \text{N} \][/tex]
2. Sum the components of the forces:
The net force components in the \( x \) (East-West) direction:
[tex]\[ F_{net_x} = F_{1_x} + F_{2_x} = 62.12109111969686 + 21.412179403483464 = 83.53327052318032 \, \text{N} \][/tex]
The net force components in the \( y \) (North-South) direction:
[tex]\[ F_{net_y} = F_{1_y} + F_{2_y} = 27.658091729154414 + 23.780634415276616 = 51.43872614443103 \, \text{N} \][/tex]
3. Calculate the magnitude of the net force:
[tex]\[ F_{net} = \sqrt{F_{net_x}^2 + F_{net_y}^2} = \sqrt{(83.53327052318032)^2 + (51.43872614443103)^2} = 98.10071269700643 \, \text{N} \][/tex]
4. Determine the direction of the net force:
[tex]\[ \text{angle}_{net} = \tan^{-2}( \frac{F_{net_y}}{F_{net_x}}) = \tan^{-2}( \frac{51.43872614443103}{83.53327052318032}) = 31.624233475557737^\circ \, \text{North of East} \][/tex]
5. Calculate the acceleration using Newton's second law:
[tex]\[ a = \frac{F_{net}}{m} = \frac{98.10071269700643}{16} = 6.131294543562902 \, \text{m/s}^2 \][/tex]
So, the magnitude and direction of the object's acceleration are:
[tex]\[ a = 6.1 \, \text{m/s}^2, 31.6^\circ \text{North of East} \][/tex]
Checking the provided options:
- A) \(1.6 \, \text{m/s}^2, 5.5^\circ \text{North of East}\)
- B) \(2.4 \, \text{m/s}^2, 34^\circ \text{North of East}\)
- C) \(3.6 \, \text{m/s}^2, 5.5^\circ \text{North of West}\)
- D) \(4.1 \, \text{m/s}^2, 52^\circ \text{North of East}\)
Since our calculated results do not match any of the provided options exactly, it suggests there may be a miscalculation or misunderstanding in the setup. However, following through with the given conditions, option A is the closest to the correct pattern, but still not the exact match.
Let's break down the process step by step:
1. Determine the components of each force:
\( F_1 = 68 \, \text{N} \) at \( 24^\circ \) North of East.
[tex]\[ F_{1_x} = F_1 \cos(24^\circ) = 62.12109111969686 \, \text{N} \][/tex]
[tex]\[ F_{1_y} = F_1 \sin(24^\circ) = 27.658091729154414 \, \text{N} \][/tex]
\( F_2 = 32 \, \text{N} \) at \( 48^\circ \) North of West.
[tex]\[ F_{2_x} = F_2 \cos(48^\circ) = 21.412179403483464 \, \text{N} \][/tex]
[tex]\[ F_{2_y} = F_2 \sin(48^\circ) = 23.780634415276616 \, \text{N} \][/tex]
2. Sum the components of the forces:
The net force components in the \( x \) (East-West) direction:
[tex]\[ F_{net_x} = F_{1_x} + F_{2_x} = 62.12109111969686 + 21.412179403483464 = 83.53327052318032 \, \text{N} \][/tex]
The net force components in the \( y \) (North-South) direction:
[tex]\[ F_{net_y} = F_{1_y} + F_{2_y} = 27.658091729154414 + 23.780634415276616 = 51.43872614443103 \, \text{N} \][/tex]
3. Calculate the magnitude of the net force:
[tex]\[ F_{net} = \sqrt{F_{net_x}^2 + F_{net_y}^2} = \sqrt{(83.53327052318032)^2 + (51.43872614443103)^2} = 98.10071269700643 \, \text{N} \][/tex]
4. Determine the direction of the net force:
[tex]\[ \text{angle}_{net} = \tan^{-2}( \frac{F_{net_y}}{F_{net_x}}) = \tan^{-2}( \frac{51.43872614443103}{83.53327052318032}) = 31.624233475557737^\circ \, \text{North of East} \][/tex]
5. Calculate the acceleration using Newton's second law:
[tex]\[ a = \frac{F_{net}}{m} = \frac{98.10071269700643}{16} = 6.131294543562902 \, \text{m/s}^2 \][/tex]
So, the magnitude and direction of the object's acceleration are:
[tex]\[ a = 6.1 \, \text{m/s}^2, 31.6^\circ \text{North of East} \][/tex]
Checking the provided options:
- A) \(1.6 \, \text{m/s}^2, 5.5^\circ \text{North of East}\)
- B) \(2.4 \, \text{m/s}^2, 34^\circ \text{North of East}\)
- C) \(3.6 \, \text{m/s}^2, 5.5^\circ \text{North of West}\)
- D) \(4.1 \, \text{m/s}^2, 52^\circ \text{North of East}\)
Since our calculated results do not match any of the provided options exactly, it suggests there may be a miscalculation or misunderstanding in the setup. However, following through with the given conditions, option A is the closest to the correct pattern, but still not the exact match.
