Question

A body of weight 12KN is lying on horizontal plane for which
-0.70. Determine normal reaction, limiting frictional force,
horizontal force required to move it.​

Answer 1

Answer:

A body of weight 12KN is lying on horizontal plane for which

-0.70. Determine normal reaction, limiting frictional force,

horizontal force required to move it.

Answer 2

Answer:

Normal reaction = R= 12 KN

Limiting Force = F = 8.4 KN

Horizontal Force = P = 8.4 KN

Angle of friction = 35°

Step-by-step explanation:

As shown in the figure,

Firstly we calculate

1)sum Fx = 0 -----Equilibrium

P -F =0

As F = u . R

P - u.R =0

P = u.R

p = 0.70 .R ---- eqn 1

2)sum Fy =0

R- w =0

R = W

R = 12 ----- put in eqn 1

we get,

P = 0.70× 12

P= 8.4 KN

Limiting force =

F = u.R

F = 8.4 KN

Angle of friction,

angle = tan ^-1( F/R)

angle = tan ^-1 (8.4/ 12)

Angle= 34.99°

Angle of friction is 35°.