To find the frictional force acting on the block, we need to consider the forces acting on the block. Let's break this down step by step.
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Identify the forces acting on the block:
- Weight component parallel to the incline (\( W_{\parallel} \)): This is given as 54.26 N.
- Normal force (\( N \)): This acts perpendicular to the inclined surface.
- Frictional force (\( F_f \)): This opposes the motion of the block.
- Applied force (\( F_a \)): This is the force required to move the block across the horizontal plane, which is given as 120 N.
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Calculate the frictional force when the block is on a horizontal plane:
- When the block is on a horizontal plane, the normal force \( N \) equals the weight of the block \( W \) (which we need to calculate first).
The total weight \( W \) can be found from the parallel weight component since: \[ W_{\parallel} = W \sin(\theta) \] Given \( W_{\parallel} = 54.26 , \text{N} \) and \( \theta = 18^\circ \): \[ W = \frac{W_{\parallel}}{\sin(\theta)} = \frac{54.26}{\sin(18^\circ)} \]
The sine of 18 degrees is approximately 0.309. Thus: \[ W \approx \frac{54.26}{0.309} \approx 175.5 , \text{N} \]
Now, when the block is moving on a horizontal plane, the normal force \( N \) is equal to the weight \( W \): \[ N = W = 175.5 , \text{N} \]
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Calculate the frictional force \( F_f \): The force required to move the block horizontally is given as 120 N. This force must overcome both the frictional force and any other forces acting on the block. In our case, since the incline is not relevant on the horizontal plane, we can set up the equation: \[ F_a = F_f \] Therefore, the frictional force \( F_f \) acting on the block when it is moving horizontally is equal to the applied force: \[ F_f = 120 , \text{N} \]
Thus, the frictional force acting on the block is \( 120 , \text{N} \).