A stepladder of negligible weight is constructed as shown in the figure below, w

Question

A stepladder of negligible weight is constructed as shown in the figure below, with AC -BC--4.40 m. A painter of mass m -75.0 kg stands on the ladder d-3.00 m from the bottom. Assuming the floor is frictionless, find the following. (Suggestion: Treat the ladder as a single object, but also treat each half of the ladder separately.) (a) the tension in the horizontal bar DE connecting the two halves of the ladder 151.3 It may be easier to work out the normal forces (part b) before solving this part. N (b) the normal forces at A and B at A at B (c) the components of the reaction force at the single hinge C that the left half of the ladder exerts on the right half rightward component upward component

Explanation / Answer

theta = angle between ground and ladder

cos theta = (L/4)/L = 1/4

theta = arccos (1/4) = 75.5 deg

In left half of ladder

Now using Force balacne in Horizontal direction:

Fx_net = 0 = T - Rx

T = Tension in the DE

Rx = resction force at the hinge C in horizontal direction

So, T = Rx

Using Force balance in Vertical direction:

Fy_net = 0 = Ry + Na - W

W = Weight of painter = 75*9.8 = 735 N

Na = Normal force at A

Ry + Na = 735 N

Torque Balance in left half about top of the ladder

Net torque = 0 = W*(d - L)*cos 75.5 deg + T*(L/2)*sin 75.5 deg - Na*L*cos 75.5 deg

735*(4.40 - 3)*cos 75.5 deg + T*2.2*sin 75.5 deg = Na*4.4*cos 75.5 deg

257.64 + T*2.13 = Na*1.1

Now In the right half of ladder

Horizontal force balance will give

T = Rx

Vertical Force balance will be

Fy_net = 0 = Nb - Ry

Nb = Ry

Torque balance about the top will be

Nb*L*cos 75.5 deg - T*(L/2)*sin 75.5 deg = 0

Nb*1.1 = T*2.13

Now We have these five equations:

T = Rx

Ry + Na = 735

257.64 + T*2.13 = Na*1.1

Nb = Ry

Nb*1.1 = T*2.13

Solving above five equations: (reduce them into 3 equation by replacing Rx with T and Ry with Nb)

Nb + Na = 735

257.64 + T*2.13 = Na*1.1

Nb*1.1 = T*2.13

Now Solving above three equations

T = 129.31 N (part A)

Na = 484.61 N & Nb = 250.39 N (Part B)

Rx = 129.31 N & Ry = 250.39 N (Part C)

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