# Video: Analysis of the Limiting Equilibrium of a Body Resting on a Rough Horizontal Plane and Attached by Two Strings

A box weighing 𝑤 kg-wt rests on a rough horizontal plane. The coefficient of static friction between the box and the plane is 1/3. Two horizontal strings, which make an angle of 90° to one another, are pulling on the box. The tension in each string is 40 kg-wt and 42 kg-wt, respectively. Given that the box is on the point of moving, find its weight 𝑤.

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### Video Transcript

A box weighing 𝑤 kilograms-weight rests on a rough horizontal plane. The coefficient of static friction between the box and the plane is one-third. Two horizontal strings, which make an angle of 90 degrees to one another, are pulling on the box. The tension in each string is 40 kilograms-weight and 42 kilograms-weight, respectively. Given that the box is on the point of moving, find its weight 𝑤.

Okay, so in this situation we have this box, and let’s imagine we’re looking down on it from above. So there’s the box and then there’s the surface it rests on beneath it. We’re told that attached to the box, there are these two horizontal strings 90 degrees apart from one another. They pull on the box, one with a force of 42 kilograms-weight and the other with 40 kilograms-weight. And under the influence of these forces, the box is just on the point of moving. Knowing that the coefficient of friction between the box and the plane, we’ll call it 𝜇, is one-third, we want to solve for the weight of this box 𝑤 in units of kilograms-weight.

Clearing some space to work, we can start out by noting that in this horizontal plane our two strings create a combined or a net force that would look like this acting on the box. We’ll call this force 𝐹 sub P since it’s the total pulling force on the box. Now, interestingly, 𝐹 sub P doesn’t make this box move because there’s an equal and opposite force, the friction force, involved. Thinking in terms of force magnitudes, we can write that 𝐹 the friction force is equal to the pull force. The pulling force 𝐹 sub P is something we can solve for. That’s because we know its orthogonal components, the pulling forces of each string independently.

Thinking of 𝐹 sub P has the hypotenuse of a right triangle, the Pythagorean theorem tells us that 𝐹 sub P would equal the square root of 42 squared plus 40 squared, where here we’ve left out the units. This comes out to exactly 58. So 𝐹 sub P is 58 kilograms-weight. We now know the total frictional force involved to keep our box in place. In general, the frictional force that acts on a body is equal to the coefficient of friction 𝜇 multiplied by the reaction force the body experiences. If we were to look at a side-on view of our body, that reaction force would look like this. In this scenario, this force is equal in magnitude but opposite in direction to the weight force acting on our body.

Getting back then to our equation for the frictional force, not only can we replace capital 𝐹 with 𝜇 times 𝑅, but we can go a step further and replace 𝑅 by 𝑤, the weight force acting on our box. This is because our box is in equilibrium, so the vertical forces acting on it are equal in magnitude. So then, dividing both sides of this equation by the coefficient of friction 𝜇, we find that 𝑤 equals 58, which, remember, has units of kilograms-weight, divided by 𝜇, which we were told is equal to one-third. 𝑤 then equals three times 58 or 174. And this weight is in units of kilograms-weight. The weight of our box then is 174 kilograms-weight.