Question

A four bar link mechanism with the following dimensions is acted upon by a force \(80 \mathrm{~N} 150^{\circ}\) on the link DC [as shown in the Figure below]: \(\mathrm{AD}=500 \mathrm{~mm}, \mathrm{AB}=400 \mathrm{~mm}, \mathrm{BC}=1000 \mathrm{~mm}, \mathrm{DC}\)

\(=750 \mathrm{~mm}, \mathrm{DE}=350 \mathrm{~mm}\) Determine the input torque \(\mathrm{T}\) on the link \(\mathrm{AB}\) for the static equilibrium of the mechanism for the given configuration.

Show all your calculations, drawing and septs to solve this problem

Answer 1

Given :

AD = 500mm, AB = 400 mm< BC = 1000 mm, DC = 750 mm

LINK DC has Force = 80 N at 150 degrees

Here Scale for force diagram is taken as 1 cm = 20 N

Refer the image.

Step 1 : The free body diagram of all the links is drawn.

Step 2 : Draw the force traiangle and find the force which comes out to be = 34N

Step 3: Find height h which comes out to be 4 cm

Step 4 : Find torque = Force * height * scale = 34 * 4 * 80 = 10880 N mm

DCE + 750 mm GIVEN AD-500mm, Ab- 400mm, BC. Joon mum Link oo & Porce-80N at 150° To find Porque link AB С. SCALE SON - 4cm 20N Icm 3 150° F =80N 120 A Free Body for AB FAB BC CD F34 F32 B F3 4 = 1.7 cm x 20 34N h - 4cm 5/50 Force = 80N F-80N Fiat Force A & Here Torque AB= force x Radin x Scale 34 X 4 X80 = dage Non Hà [O, 880 Nam