Question

A man pulls a loaded trolley of mass 70 kg along a horizontal surface at constant velocity as shown in figure. The coefficient of kinetic friction mu i between the tyres of the trolley and the road is 0.1 and the angle made by the rope with the horizontalis 45° Calculate the tension in the rope.​

Answer 1

Answer:

given = diagram .

            mass = 70Kg

            μ  =  0.1

            angle =  45°

to find =  tension in the rope.​

solution = Tcos45° - fk= 0

                fk =  Tcos45° = T X 1/ √2 = 0.7072T

but,          fk = μkR

              μR = 0.7072T

               R = 0.7072/μk

   

TSin45° + R = mg

T Sin45° + R -mg = 0

T / √2 + R - mg = 0

0.7072 T + R -mg = 0

0.7072T + 0.07072 /μk - mg = 0

T(0.7072μg + 0.7072/μg -mg = 0

T =  mgμk/0.7072μg = 0.7072

T = 70 X 9.8 X 0.1

T =  616N