Force, Weight, And Reaction Angle On A Horizontal Plane

Let's break down this physics problem step by step. We're dealing with a body resting on a rough horizontal plane, subjected to a horizontal force, and we need to figure out the angle between the resultant reaction force and the horizontal force. This involves understanding concepts like weight, normal reaction, friction, and resultant forces. Ready to dive in, guys?

Understanding the Forces at Play

First off, let's identify all the forces acting on the body:

• Weight (W): This is the force due to gravity, acting vertically downwards. We're given that the weight W = 15 N.
• Normal Reaction (R): This is the force exerted by the horizontal plane on the body, acting vertically upwards, perpendicular to the surface. In equilibrium (when the body is not accelerating vertically), the normal reaction R is equal in magnitude to the weight W. So, R = 15 N.
• Applied Horizontal Force (F): This is the external force we're applying, which tends to move the body horizontally.
• Frictional Force (f): This force opposes the motion (or the tendency of motion) of the body along the surface. It acts horizontally and is directly related to the normal reaction and the coefficient of static friction (μ) between the body and the plane. The frictional force f ≤ μR.

Resultant Reaction

The resultant reaction is the vector sum of the normal reaction (R) and the frictional force (f). Let's call this resultant reaction S. The normal reaction acts vertically upwards, and the frictional force acts horizontally, opposing the applied force F. Therefore, these two forces are perpendicular to each other. We can express S as a vector sum:

$\vec{S} = \vec{R} + \vec{f}$

The magnitude of S can be found using the Pythagorean theorem:

$S = \sqrt{R^2 + f^2}$

Since R = 15 N, we have:

$S = \sqrt{15^2 + f^2}$

Angle Between Resultant Reaction and Horizontal Force

The problem asks for the angle θ between the resultant reaction S and the horizontal force F. To visualize this, imagine a right-angled triangle where:

• The vertical side is the normal reaction R.
• The horizontal side is the frictional force f.
• The hypotenuse is the resultant reaction S.

The angle between S and the vertical (i.e., R) is given by:

$\tan^{-1}(\frac{f}{R})$

However, we need the angle between S and the horizontal force F. Since f opposes F, we can consider the angle between S and f. Let’s call the angle between S and the horizontal frictional force α. Then:

$\tan(\alpha) = \frac{R}{f} = \frac{15}{f}$

So,

$\alpha = \tan^{-1}(\frac{15}{f})$

Now, here's the tricky part. The question asks for the angle θ between the resultant reaction S and the applied horizontal force F. Since the frictional force f directly opposes F, we can say that the angle between S and F is the same as the angle between S and f (i.e., α). Therefore:

$\theta = \alpha = \tan^{-1}(\frac{15}{f})$

The angle θ depends on the magnitude of the frictional force f. The maximum value of f is μR, where μ is the coefficient of static friction. Therefore, the minimum value of θ occurs when f is maximum:

$\theta_{min} = \tan^{-1}(\frac{15}{\mu R}) = \tan^{-1}(\frac{15}{15\mu}) = \tan^{-1}(\frac{1}{\mu})$

Without knowing the exact value of F or μ, we can only express θ in terms of f or μ.

Discussion and Possible Scenarios

Let's consider different scenarios to understand this better:

1. If the body is just about to move: In this case, the frictional force f is at its maximum value, f = μR. Then, θ = $\tan^{-1}(\frac{1}{\mu})$. The angle depends entirely on the coefficient of static friction.
2. If there is no friction (μ = 0): This is an idealized situation, but if there were no friction, then f = 0, and the resultant reaction S would be equal to the normal reaction R. In this case, the angle θ would be 90 degrees, as S would be vertical, and F is horizontal.
3. If the body is in equilibrium and not about to move: Here, the frictional force f is equal to the applied force F, and f < μR. The angle θ can be any value between $\tan^{-1}(\frac{1}{\mu})$ and 90 degrees, depending on the magnitude of F.

Example Calculation

Suppose the coefficient of static friction μ = 0.5. This means the maximum frictional force is f = 0.5 * 15 = 7.5 N. If the body is just about to move, then:

$\theta = \tan^{-1}(\frac{15}{7.5}) = \tan^{-1}(2) \approx 63.43^{\circ}$

This tells us that the resultant reaction S is at an angle of approximately 63.43 degrees with respect to the horizontal force F when the body is on the verge of moving.

Final Thoughts

To summarize, the angle θ between the resultant reaction and the horizontal force is given by $\theta = \tan^{-1}(\frac{15}{f})$, where f is the frictional force. The value of θ depends on the magnitude of f, which is influenced by the applied force F and the coefficient of static friction μ. Understanding these relationships helps in analyzing the forces acting on a body on a rough horizontal plane. Keep experimenting and exploring, physics nerds!