# Question Video: Finding the Reaction of a Smooth Inclined Plane on a Body with a Force Acting on the Body

A body of mass 1.3 kg was placed on a smooth plane inclined at 60° to the horizontal. A force of 62 N was acting on the body along the line of greatest slope of the plane in the upward direction. Find the magnitude of the reaction of the plane. Take 𝑔 = 9.8 m/s².

03:10

### Video Transcript

A body of mass 1.3 kilograms was placed on a smooth plane inclined at 60 degrees to the horizontal. A force of 62 newtons was acting on the body along the line of greatest slope of the plane in the upward direction. Find the magnitude of the reaction of the plane. Take 𝑔 to be equal to 9.8 meters per square second.

Let’s begin by sketching out the diagram. Remember, this diagram doesn’t need to be to scale, but it should be roughly in proportion. We have a smooth plane inclined at 60 degrees to the horizontal. A body with mass 1.3 kilograms rests on that plane, and this means it exerts a downward force on the plane itself. That downwards force, according to Newton’s second law, is equal to its mass times acceleration due to gravity. Since the mass is 1.3 kilograms, that’s 1.3𝑔. Then there’s a force of 62 newtons acting on the body along the line of greatest slope in the upward direction, so that’s shown. Newton’s third law of motion tells us that there is a normal reaction force of the plane on the body that acts away from the plane at an angle of 90 degrees as shown.

We’re trying to find the magnitude of that reaction force 𝑅. So let’s set up and solve some equations. We’re going to resolve forces perpendicular to the plane. And so to do so, we’re going to need to work out the component of the weight force of the body on the plane that acts in this direction. And so we add in a right-angled triangle as shown. The included angle in that triangle is 60 degrees. And we’re trying to find the magnitude of the length of the side adjacent to that. Let’s label that 𝑥, and this will be in newtons.

We know that the cosine ratio tells us that cos 𝜃 is adjacent over hypotenuse. Now we’re looking to find the adjacent, and we know the length of the hypotenuse. It’s 1.3𝑔. So we substitute everything we know into the formula, and we get cos of 60 equals 𝑥 over 1.3𝑔. cos of 60 degrees is 0.5. And then we can solve for 𝑥 by multiplying through by 1.3𝑔, and so we get 𝑥 is equal to 0.5 times 1.3𝑔. We were told, though, that 𝑔 is 9.8 meters per square second, so we can rewrite this as 0.5 times 1.3 times 9.8. And that gives us a value of 6.37. So the component of the weight that acts perpendicular to the plane is 6.37.

We know the object is at rest. It’s in equilibrium, so the sum of the forces in this direction have to be zero. If we take upwards and away from the plane to be positive, in other words, the direction that 𝑅 is moving, the sum of the forces acting in this direction is 𝑅 minus 𝑥. That’s 𝑅 minus 6.37. So, our equation becomes 𝑅 minus 6.37 equals zero. We solve this equation for 𝑅 by adding 6.37 to both sides. And when we do, we get 𝑅 is equal to 6.37. And so the magnitude of the reaction on the plane 𝑅 is 6.37 newtons.