A body of mass 15 kg is on a rough plane inclined at an angle of 25° to the horizontal. The coefficient of friction between the body and the plane is 0·4. The body is held at rest by means of a light string, which is parallel to a line of greatest slope of the plane. The magnitude of the tension in the string is T N. Find the greatest and the least possible values of T, giving your answers correctto two decimal places.
1 Answer | Add Yours
The body of mass 15 kg is on a rough plane inclined at an angle of 25° to the horizontal. The coefficient of friction between the body and the plane is 0.4. It is held stationary by means of a string that is parallel to the line of greatest slope of the plane.
The gravitational force of attraction due to the Earth exerts a force equal to 15*9.8 = 147 N in a vertically downwards direction. The component of this force in a direction perpendicular to the plane is 147*sin 65 `~~` 133.22 N and the component in a direction parallel to the plane is 147*cos 65 `~~` 61.12. The force of frictional between the plane and the body is equal to 133.22*0.4 = 53.29 N
This gives the net force acting on the body when it is at rest as 61.12 - 53.29 = 7.83 N. The minimum tension in the string is 7.83 N, a lower value would allow the body to move downwards. The maximum tension in the string is equal to 61.12 + 53.39 = 114.41 N. If the tension were raised above this value, the body would move up the plane.
We’ve answered 334,087 questions. We can answer yours, too.Ask a question