Question

As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m/s²) (Hint: Consider the equilibrium of each side of the ladder separately.)

Answer 1

Given:the suspended mass is 40 kg, the length of the rope is 0.5 m, the length of BA and CA is 1.6 m, and the length of BF is 1.2 m.

The free body diagram of the ladder system is shown below:

Here, N B is the normal reaction force at point B, N C is the normal reaction force at point C, and the suspended weight is W.

Take the net moment about point A.

N B ×BI+T×AH−W×FG−T×AH− N C ×CI=0 N B ×BI−W×FG− N C ×CI=0 …… (1)

As triangle ABI and triangle ACI are similar, therefore,

BI=CI

Substitute BI for CI in the equation (1).

N B ×BI−W×FG− N C ×BI=0 ( N B − N C )BI=W×FG N B − N C = W×FG BI …… (2)

As triangle AFG and triangle ABI are similar, therefore,

FG BI = AF AB FG BI = AB−BF AB = 1.6−1.2 1.6 =0.25

Substitute the value of FG BI in the equation (2).

N B − N C =W×0.25 N B − N C =0.25W ….. (3)

The equation for the weight of the suspended mass is,

W=mg

Here, m is the suspended mass and g is the gravitational acceleration.

Substitute the given values in the above equation.

W=40×9.8 =392 N

The equilibrium equation of forces in the vertical direction is,

N B + N C =W …… (4)

From equations (3) and (4),

2 N B = 5W 4 N B = 5W 8

Substitute the value of W in the above equation.

N B = 5mg 8

Substitute the given values in the above equation.

N B = 5×40×9.8 8 =245 N

Substitute 5W 8 for N B in the equation (4).

5W 8 + N C =W N C = 3W 8

Substitute the value of W in the above equation.

N C = 3mg 8

Substitute the given values in the above equation.

N C = 3×40×9.8 8 =147 N

Under the balanced condition, the net torque about point A due to the force acting on the part AB should balance the torque developed due to the force acting on the part AC.

For side AB, take the net moment about point A.

− N B ×BI+W×FG−T×AH=0

Substitute the given values in the above equation.

−245×BI+392×FG−T×AH=0 …… (5)

From the triangle AHD,

( AD ) 2 = ( AH ) 2 + ( DH ) 2 ( AB 2 ) 2 = ( AH ) 2 + ( DE 2 ) 2 AH= 1 2 ( ( AB ) 2 − ( DE ) 2 )

Substitute the values in the above equation.

AH= 1 2 ( ( 1.6 ) 2 − ( 0.5 ) 2 ) =0.76 m

As triangle ADH and triangle ABI are similar, therefore,

AH DH = AI BI

Substitute the given values in the above equation.

0.76 ( 0.5 2 ) = ( AB ) 2 − ( BI ) 2 BI 3.04 2 = ( 1.6 ) 2 − ( BI ) 2 ( BI ) 2 9.2416 ( BI ) 2 =2.56− ( BI ) 2 BI=0.5 m

The ratio FG BI is,

FG BI =0.25

Substitute the values in the above equation.

FG 0.5 =0.25 FG=0.125 m

Substitute the values for BI, FG, and AH in the equation (5).

−245×0.5+392×0.125+T×0.76=0 T=96.71 N ≈98 N

Thus, tension in the rope is 98 N and the forces exerted by the floor on the ladder are N B =245 N and N C =147 N respectively.