The figure below shows a toolbox of weight 300N on a rough horizontal plane having a coefficient of friction as 0.2. Calculate the magnitude of the force which can pull the box while acting at an angle of 25 degrees with the horizontal.

Question

Asked on July 7, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-07-07T18:04:20.581Z

Step 1: Identify the given values and set up the force equilibrium equations. The weight of the toolbox, W = 300 \, N (taken from the diagram, as it is more clearly indicated than the text). The coefficient of friction, = 0.2. The angle of the pulling force with the horizontal, = 25^. Let P be the magnitude of the pulling force and R be the normal force exerted by the surface on the toolbox. The friction force f_f opposes the motion. The friction force is given by: f_f = R For vertical equilibrium (no vertical acceleration): The sum of upward forces equals the sum of downward forces. The upward forces are the normal force R and the vertical component of the pulling force P . The downward force is the weight W. R + P = W R = W - P (1) For horizontal equilibrium (constant velocity or on the verge of motion): The sum of forces in the direction of motion equals the sum of forces opposing motion. The force in the direction of motion is the horizontal component of the pulling force P . The force opposing motion is the friction force f_f. P = f_f (2) Step 2: Substitute the friction force into the horizontal equilibrium equation. Substitute f_f = R into equation (2): P = R Step 3: Substitute the expression for the normal force R from equation (1). P = (W - P ) P = W - P Step 4: Rearrange the equation to solve for P. Move all terms containing P to one side: P + P = W Factor out P: P ( + ) = W Solve for P: P = ( W)/( + ) Step 5: Substitute the numerical values and calculate P. Given W = 300 \, N, = 0.2, and = 25^. First, calculate 25^ and 25^: 25^ ≈ 0.9063 25^ ≈ 0.4226 Now substitute these values into the equation for P: P = (0.2 × 300)/(0.9063 + 0.2 × 0.4226) P = (60)/(0.9063 + 0.08452) P = (60)/(0.99082) P ≈ 60.5555 \, N Rounding to three significant figures: P ≈ 60.6 \, N The magnitude of the force required to pull the box is 60.6 N. Send me the next one 📸

Accepted Answer · 2026-07-07T18:04:20.581Z

Step 1: Identify the given values and set up the force equilibrium equations. The weight of the toolbox, W = 300 \, N (taken from the diagram, as it is more clearly indicated than the text). The coefficient of friction, = 0.2. The angle of the pulling force with the horizontal, = 25^. Let P be the magnitude of the pulling force and R be the normal force exerted by the surface on the toolbox. The friction force f_f opposes the motion. The friction force is given by: f_f = R For vertical equilibrium (no vertical acceleration): The sum of upward forces equals the sum of downward forces. The upward forces are the normal force R and the vertical component of the pulling force P . The downward force is the weight W. R + P = W R = W - P (1) For horizontal equilibrium (constant velocity or on the verge of motion): The sum of forces in the direction of motion equals the sum of forces opposing motion. The force in the direction of motion is the horizontal component of the pulling force P . The force opposing motion is the friction force f_f. P = f_f (2) Step 2: Substitute the friction force into the horizontal equilibrium equation. Substitute f_f = R into equation (2): P = R Step 3: Substitute the expression for the normal force R from equation (1). P = (W - P ) P = W - P Step 4: Rearrange the equation to solve for P. Move all terms containing P to one side: P + P = W Factor out P: P ( + ) = W Solve for P: P = ( W)/( + ) Step 5: Substitute the numerical values and calculate P. Given W = 300 \, N, = 0.2, and = 25^. First, calculate 25^ and 25^: 25^ ≈ 0.9063 25^ ≈ 0.4226 Now substitute these values into the equation for P: P = (0.2 × 300)/(0.9063 + 0.2 × 0.4226) P = (60)/(0.9063 + 0.08452) P = (60)/(0.99082) P ≈ 60.5555 \, N Rounding to three significant figures: P ≈ 60.6 \, N The magnitude of the force required to pull the box is 60.6 N. Send me the next one 📸

The figure below shows a toolbox of weight 300N on a rough horizontal plane having a coefficient of friction as 0.2. Calculate the magnitude of the force which can pull the box while acting at an angle of 25 degrees with the horizontal.

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The figure below shows a toolbox of weight 300N on a rough horizontal plane having a coefficient of friction as 0.2. Calculate the magnitude of the force which can pull the box while acting at an angle of 25 degrees with the horizontal.

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