Answer :
To determine the unit vector in the direction of the friction force acting on the block, we must first understand that friction opposes the relative motion between the block and the plank.
Given:
- The block is moving along the [tex]\(y\)[/tex]-axis with a velocity of [tex]\(3 \, \text{m/s}\)[/tex].
- The plank is moving along the [tex]\(x\)[/tex]-axis with a velocity of [tex]\(4 \, \text{m/s}\)[/tex].
Let's find the relative velocity of the block with respect to the plank:
Since the plank is moving along the [tex]\(x\)[/tex]-axis and the block along the [tex]\(y\)[/tex]-axis, these motions are perpendicular to each other. Hence, the relative velocity of the block with respect to the plank can be represented as a vector:
[tex]\[ \mathbf{v}_{\text{relative}} = 3 \hat{j} - 4 \hat{i} \][/tex]
Here, [tex]\(3 \hat{j}\)[/tex] is the velocity of the block along the [tex]\(y\)[/tex]-axis and [tex]\(-4 \hat{i}\)[/tex] is the velocity of the block relative to the plank along the negative [tex]\(x\)[/tex]-axis (as we are considering the relative motion), hence:
[tex]\[ \mathbf{v}_{\text{relative}} = -4 \hat{i} + 3 \hat{j} \][/tex]
The friction force will act in the opposite direction to the relative velocity. Therefore, the direction of the friction force will be opposite to [tex]\(\mathbf{v}_{\text{relative}}\)[/tex]:
[tex]\[ \mathbf{f}_{\text{friction}} = -( -4 \hat{i} + 3 \hat{j}) = 4 \hat{i} - 3 \hat{j} \][/tex]
To find the unit vector in the direction of the friction force, we first need to determine the magnitude of [tex]\(\mathbf{f}_{\text{friction}}\)[/tex]:
[tex]\[ |\mathbf{f}_{\text{friction}}| = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \][/tex]
Now, the unit vector in the direction of the friction force is calculated by dividing [tex]\(\mathbf{f}_{\text{friction}}\)[/tex] by its magnitude:
[tex]\[ \hat{f}_{\text{friction}} = \frac{4 \hat{i} - 3 \hat{j}}{5} \][/tex]
Hence, the unit vector in the direction of the friction force acting on the block is given by:
[tex]\[ \boxed{\frac{4 \hat{i} - 3 \hat{j}}{5}} \][/tex]
Given:
- The block is moving along the [tex]\(y\)[/tex]-axis with a velocity of [tex]\(3 \, \text{m/s}\)[/tex].
- The plank is moving along the [tex]\(x\)[/tex]-axis with a velocity of [tex]\(4 \, \text{m/s}\)[/tex].
Let's find the relative velocity of the block with respect to the plank:
Since the plank is moving along the [tex]\(x\)[/tex]-axis and the block along the [tex]\(y\)[/tex]-axis, these motions are perpendicular to each other. Hence, the relative velocity of the block with respect to the plank can be represented as a vector:
[tex]\[ \mathbf{v}_{\text{relative}} = 3 \hat{j} - 4 \hat{i} \][/tex]
Here, [tex]\(3 \hat{j}\)[/tex] is the velocity of the block along the [tex]\(y\)[/tex]-axis and [tex]\(-4 \hat{i}\)[/tex] is the velocity of the block relative to the plank along the negative [tex]\(x\)[/tex]-axis (as we are considering the relative motion), hence:
[tex]\[ \mathbf{v}_{\text{relative}} = -4 \hat{i} + 3 \hat{j} \][/tex]
The friction force will act in the opposite direction to the relative velocity. Therefore, the direction of the friction force will be opposite to [tex]\(\mathbf{v}_{\text{relative}}\)[/tex]:
[tex]\[ \mathbf{f}_{\text{friction}} = -( -4 \hat{i} + 3 \hat{j}) = 4 \hat{i} - 3 \hat{j} \][/tex]
To find the unit vector in the direction of the friction force, we first need to determine the magnitude of [tex]\(\mathbf{f}_{\text{friction}}\)[/tex]:
[tex]\[ |\mathbf{f}_{\text{friction}}| = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \][/tex]
Now, the unit vector in the direction of the friction force is calculated by dividing [tex]\(\mathbf{f}_{\text{friction}}\)[/tex] by its magnitude:
[tex]\[ \hat{f}_{\text{friction}} = \frac{4 \hat{i} - 3 \hat{j}}{5} \][/tex]
Hence, the unit vector in the direction of the friction force acting on the block is given by:
[tex]\[ \boxed{\frac{4 \hat{i} - 3 \hat{j}}{5}} \][/tex]
