Question Video: Find the Acceleration of a Body Moving on an Inclined Plane with a Force Acting on It | Nagwa Question Video: Find the Acceleration of a Body Moving on an Inclined Plane with a Force Acting on It | Nagwa

Question Video: Find the Acceleration of a Body Moving on an Inclined Plane with a Force Acting on It Mathematics

A body of mass 484 g was placed on a smooth plane inclined at an angle πœƒ to the horizontal, where sin πœƒ = 3/5. Given that a force of 484 g-wt was acting on the body parallel to the line of greatest slope up the plane, find the acceleration of motion. Consider the acceleration due to gravity to be 9.8 m/sΒ².

04:11

Video Transcript

A body of mass 484 grams was placed on a smooth plane inclined at an angle πœƒ to the horizontal, where sin πœƒ is equal to three-fifths. Given that a force of 484 gram-weight was acting on the body parallel to the line of greatest slope up the plane, find the acceleration of motion. Consider the acceleration due to gravity to be 9.8 meters per second squared.

We will begin by sketching a diagram to model the scenario. We are told that the body has a mass of 484 grams. And since there are 1000 gram in a kilogram, this is equal to 0.484 kilograms. The body exerts a force vertically downwards equal to its weight, which is equal to 0.484 multiplied by gravity 𝑔. We are told that this is equal to 9.8 meters per second squared. This means that there is a force of 4.7432 newtons acting vertically downwards. There will be a normal reaction force 𝑅 acting perpendicular to the plane.

We are told that the plane is inclined at an angle πœƒ to the horizontal, where sin of πœƒ is equal to three-fifths. Since the plane is smooth, there will be no frictional force. However, there is a force of 484 gram-weight acting up the plane. As this is equal to 0.484 kilogram-weight, we can multiply this by 9.8 meters per second squared, giving us a force acting up the plane of 4.7432 newtons.

We are now in a position to resolve parallel to the plane using Newton’s second law 𝐹 equals π‘šπ‘Ž. The sum of the forces will be equal to the mass multiplied by the acceleration. We will let this acceleration up the plane be π‘Ž, and we need to calculate the component of the weight force acting down the plane. We will do this using our knowledge of right angle trigonometry. The component of the weight force acting parallel to the plane is 𝑀 sin πœƒ. And whilst it is not required in this question, the components of the weight force acting perpendicular to the plane is 𝑀 cos πœƒ.

We have two forces acting parallel to the plane. And letting the positive direction be the direction of motion, the sum of our forces is 4.7432 minus 𝑀 sin πœƒ. This is equal to the mass of 0.484 kilograms multiplied by the acceleration π‘Ž. Substituting the weight force of 4.7432 newtons and sin πœƒ equals three-fifths, we have the following equation. Typing the left-hand side into our calculator gives us 1.89728. This is equal to 0.484π‘Ž. We can then divide through by 0.484 giving us π‘Ž is equal to 3.92. The acceleration of the body is 3.92 meters per second squared up the plane.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy