Question Video: Finding the Maximum Force Required to Keep a Body in a State of Equilibrium on a Rough Inclined Plane | Nagwa Question Video: Finding the Maximum Force Required to Keep a Body in a State of Equilibrium on a Rough Inclined Plane | Nagwa

Question Video: Finding the Maximum Force Required to Keep a Body in a State of Equilibrium on a Rough Inclined Plane Mathematics

A body weighing 75 N rests on a rough plane inclined at an angle of 45° to the horizontal under the action of a horizontal force. The minimum horizontal force required to maintain the body in a state of equilibrium is 45 N. Find the maximum horizontal force that will also maintain the equilibrium.

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Video Transcript

A body weighing 75 newtons rests on a rough plane inclined at an angle of 45 degrees to the horizontal under the action of a horizontal force. The minimum horizontal force required to maintain the body in a state of equilibrium is 45 newtons. Find the maximum horizontal force that will also maintain the equilibrium.

In this question, we are looking at two different scenarios: when the force is at its minimum and maximum. In both of these situations, the body is in equilibrium. Therefore, the sum of the net forces must equal zero. We have a vertical downward force of 75 newtons equal to the weight. The plane is inclined at an angle of 45 degrees. When we have the minimum horizontal force equal to 45 newtons, the body will be on the point of sliding down the plane. This means that the frictional force will act up the plane.

We will resolve parallel and perpendicular to the plane. Before doing this, we will need to work out the components of the 75- and 45-newton forces in these directions. We do this using right angle trigonometry, giving us the four forces shown. Parallel to the plane, we have three forces. As the body is on the point of sliding down the plane, we will use this as the positive direction. This gives us the equation 75 sin 45 minus 45 cos 45 minus 𝐹 r is equal to zero. The sin of 45 degrees and cos of 45 degrees are both equal to root two over two. Adding the frictional force to both sides of this equation gives us 𝐹 r is equal to 15 root two.

There are also three forces acting perpendicular to the plane. This gives us the equation 𝑅 minus 75 cos 45 minus 45 sin 45 is equal to zero. This simplifies to 𝑅 minus 60 root two equal zero, which means that 𝑅 is equal to 60 root two. We now have the frictional force and normal reaction force in newtons. And we know that 𝐹 r equals 𝜇, the coefficient of friction, multiplied by 𝑅. 𝜇 is equal to 15 root two divided by 60 root two. This simplifies to one-quarter.

We now need to consider the case where the horizontal force is at its maximum. In this situation, most of our forces remain the same. However, the body will be on the point of moving up the plane, so the frictional force acts downwards. We need to calculate the unknown force 𝐹. Resolving parallel to the plane, we have 𝐹 cos 45 minus 75 sin 45 minus 𝐹 r is equal to zero. Simplifying this and using the fact that 𝜇 equals one-quarter gives us root two over two 𝐹 minus 75 root two over two minus one-quarter 𝑅 is equal to zero. In the perpendicular direction, we have 𝑅 minus 75 cos 45 minus 𝐹 sin 45 is equal to zero. Simplifying this, we get 𝑅 is equal to 75 root two over two plus root two over two 𝐹.

We now have two simultaneous equations that we can solve to calculate 𝐹. We will replace the value of 𝑅 in equation two into equation one. We can then divide through this whole equation by root two over two. Distributing our parentheses gives us 𝐹 minus 75 minus 75 over four minus a quarter 𝐹 is equal to zero. This in turn simplifies to three-quarters 𝐹 is equal to 375 over four. Dividing both sides of the equation by three-quarters gives us 𝐹 is equal to 125. The maximum force is equal to 125 newtons. This means that if the force is between 45 and 125 newtons inclusive, the body will remain in equilibrium.

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