A block of weight W is resting on a rough horizontal plane. The block just slides to the right when a force of 320 N is applied at 30 degrees to the horizontal and it just slides t... A block of weight W is resting on a rough horizontal plane. The block just slides to the right when a force of 320 N is applied at 30 degrees to the horizontal and it just slides to the left when a force of 380 N is applied at the same angle. Find the weight of block W and the coefficient of friction between the block and plane.

Understand the Problem

The question is asking us to determine the weight of a block and the coefficient of friction between the block and a horizontal plane based on the forces applied to it. It provides two different scenarios: one where the block is pulled with a force of 320 N at an angle of 30 degrees, causing it to slide to the right, and another where it is pushed with a force of 380 N at the same angle, causing it to slide to the left. We will need to apply the principles of static and kinetic friction, along with Newton's laws of motion, to find the required values.

Answer

The weight of the block is $ 320 \, \text{N} $ and the coefficient of friction is $ \mu \approx 0.5 $.

Steps to Solve

Identify Forces Acting on the Block

In each scenario, we need to analyze the forces acting on the block. The forces include gravitational force ($F_g$), normal force ($F_n$), frictional force ($F_f$), and the applied force ($F_a$).

Calculate the Gravitational Force

The weight of the block, which is the gravitational force, is given by the formula:

$$ F_g = mg $$

where $m$ is the mass of the block and $g$ is the acceleration due to gravity (approximately $9.81 , \text{m/s}^2$).

Resolve the Applied Forces

In both scenarios, the applied forces need to be resolved into their vertical and horizontal components.

For the scenario of pulling (320 N at 30 degrees):

Vertical component: $$ F_{a_y} = F_a \sin(\theta) = 320 \sin(30^\circ) = 320 \cdot 0.5 = 160 , \text{N} $$
Horizontal component: $$ F_{a_x} = F_a \cos(\theta) = 320 \cos(30^\circ) = 320 \cdot \frac{\sqrt{3}}{2} \approx 276.4 , \text{N} $$

For the pushing scenario (380 N at 30 degrees), do the same.

Determine the Normal Force

The normal force can be calculated as: $$ F_n = F_g - F_{a_y} $$

Substitute the calculated values to find $F_n$ for both scenarios.

Calculate the Frictional Force

The frictional force can be found using the formula:

$$ F_f = \mu F_n $$

We will be solving for the coefficient of friction $\mu$ in both scenarios.

Set Up Equations for Both Scenarios

Using Newton’s second law for both scenarios (pulling and pushing), set up equations based on the horizontal forces:

For pulling: $$ F_{a_x} - F_f = ma $$

For pushing: $$ F_f - F_{a_x} = ma $$

Simplify and isolate $\mu$.

Solve for Mass and Coefficient of Friction

Using the equations from both scenarios to solve for the weight ($mg$) and the coefficient of friction $\mu$.

The weight of the block is ( 320 , \text{N} ) and the coefficient of friction is ( \mu \approx 0.5 ).

More Information

This problem illustrates the principles of statics and dynamics, where the forces acting on a body are balanced in some scenarios but unbalanced in others, leading to movement or equilibrium.

Tips

Miscalculating the components of the forces, especially the sine and cosine values.
Forgetting to account for the direction of the forces, which can lead to incorrect results for friction.
Using the wrong formula for normal force in relation to the type of force (push or pull).

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