A rod AB rests with the end A on rough horizontal ground, such that end A is prevented from sliding to the right. End B rests on smooth vertical wall. The rod is uniform and its weight is W. If the rod is in a stationary position as shown in figure, the number of forces acting in the FBD of rod is/are

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Toppr Admin · Accepted Answer

The wall is frictionless so frictional force at $B$ will be zero.
W is weight of the road and acting along $A B$ at $A B / 2$ from $A$ .

To calculate reaction at $B$ , put moment about $A = 0$ .

$N_{B} s i n 30^{0} (A B) - W (\frac{A B}{2}) c o s 30^{0} = 0$

$⟹ N_{B} = W \frac{\sqrt{(} 3)}{2}$

A rod AB rests with the end A on rough horizontal ground and the end B against a smooth vertical wall. The rod is uniform and of weight W. If the rod is in equilibrium in the position shown in figure. Find normal reaction at B.W√32W√34W2√3W√3

The wall is frictionless so frictional force at B will be zero. W is weight of the road and acting along AB at AB/2 from A.To calculate reaction at B, put moment about A=0.NBsin300(AB)−W(AB2)cos300=0⟹NB=W√(3)2

The wall is frictionless so frictional force at $B$ will be zero.
W is weight of the road and acting along $A B$ at $A B / 2$ from $A$ .

To calculate reaction at $B$ , put moment about $A = 0$ .

$N_{B} s i n 30^{0} (A B) - W (\frac{A B}{2}) c o s 30^{0} = 0$

$⟹ N_{B} = W \frac{\sqrt{(} 3)}{2}$