A block of mass m lies on a rough plane which is inclined at an angle è to the horizontal. The coefficient of static friction between the block and the plane is ì. A force of magnitude P is now applied to the block in a horizontal direction, towards the plane.
Show that the minimum value of P which is necessary to ensure that the block remains at rest on the plane, is
P=[mg(sinè-ìcosè)]/(ìsinè+cosè)
Please help! Thanks!
Take x-axis directed along the incline downwards, and y-axis directed normaly away from the incline.
Projections of the equation of motion on the x- and y- axes are:
m•g•sinα - P•cosα – F(fr) = 0,
-m•g•cos α + N -P•sinα = 0.
F(fr) = μ •N = μ •(m•g•cos α+P•sinα).
0 = m•g•sinα - P•cosα - μ •(m•g•cos α+P•sinα) =
= m•g•sinα - P•cosα - μ •m•g•cos α+μ•P•sinα.
P = m•g•( sinα - μ•cos α)/(cos α + μ•sinα).
To determine the minimum value of P required to keep the block at rest on the inclined plane, we need to consider the forces acting on the block.
1. Draw a free-body diagram for the block.
- The weight of the block acts vertically downward and can be represented by mg, where m is the mass of the block and g is the acceleration due to gravity.
- The normal force exerted by the plane on the block acts perpendicular to the plane.
- The force of static friction acts parallel to the plane, opposing the tendency for the block to slide down the incline.
- The applied force P acts horizontally, towards the plane.
2. Decompose the weight and the normal force.
- Resolve the weight mg into two components: mg sin(è) acting parallel to the plane and mg cos(è) acting perpendicular to the plane.
- The normal force N is equal in magnitude and opposite in direction to the component mg cos(è) acting perpendicular to the plane.
3. Calculate the force of static friction.
- The maximum force of static friction can be given by the equation fs_max = ìN, where ì is the coefficient of static friction.
- Since the block is at rest, the horizontal component of the weight mg sin(è) is balanced by the applied force P and the force of static friction.
- Therefore, fs_max = mg sin(è).
- However, we are interested in determining the minimum value of P required, so we consider the force of static friction at its maximum value.
4. Set up the equation for the equilibrium of forces.
- The force of static friction fs acts in the direction opposite to the applied force P.
- The net force in the horizontal direction is given by P - fs = 0, since the block is at rest.
- Substituting the value of fs_max = mg sin(è), we have P - mg sin(è) = 0.
5. Solve the equation for P.
- Rearrange the equation to solve for P: P = mg sin(è).
- Divide both the numerator and the denominator of the equation by cos(è) to get the desired expression:
P = [mg sin(è)] / (sin(è) / cos(è)) = [mg sin(è)] / (tan(è)) = [mg(sin(è) - ì cos(è))] / (ìsin(è) + cos(è))
Therefore, the minimum value of P required to keep the block at rest on the inclined plane is P = [mg(sin(è) - ì cos(è))] / (ìsin(è) + cos(è)).