Question

A block weighing 200N is in contact with an inclined plane (Inclination = 30º). Will the block move under its own weight. Determine the minimum force applied (1) parallel (2) perpendicular to the plane to prevent the motion down the plane. What force P will be required to just cause the motion up the plane, µ = 0.25? 

Answer 1

Consider the FBD of block as shown in fig(a)

From the equilibrium condition Sum of forces perpendicular to plane = 0 

R – W cos 30° = 0; R = W cos 30° ...(i) 

Sum of forces parallel to plane = 0 

µR – W sin 30° = 0

Now body will move down only if the value of µR is less than Wsin30° 

Now, µR = 0.25 × W(0.866) = 0.2165W ...(iii) 

And Wsin30° = 0.5W ...(vi) 

Since value of (iv) is less than value of (iii) So the body will move down. 

(i) When Force acting parallel to plane as shown in fig (b) Frictional force is acting up the plane

∑V = 0; R = 0.216 W ...(v) 

∑H = 0; 

P = W sin 30° – µR 

P = 0.5 × 200 – 0.216 × 200

P = 56.7N

(ii) When Force acting perpendicular to plane as shown in fig(c) Frictional force is acting up the plane 

∑H = 0; Wsin30° – µR = 0 R = 400 ...(vii) 

∑V = 0; P + R = W cos 30° 

P = 0.866 × 200 – 400 

P = –226.79N

(iii) The force P required to just cause the motion up the plane as shown in fig(d) Frictional force is acting down the plane 

Sum of force perpendicular to plane = 0 R = W cos 30 = 172.2N ...(viii) 

Sum of force Parallel to plane = 0 

P – µR – W sin 30 = 0 

P = 0.25 × 172.2 – 200 sin 30 

P = 143.3N