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A mover has to move a heavy sofa of mass 150 kg to the second floor of the house. He uses a rope to pull the sofa up a ramp from the first to the second floor. As he pulls the sofa he makes sure that the rope is parallel to the surface of the ramp which is at 30.0° to the horizontal. If friction between the sofa and the ramp is negligible, and the sofa has an acceleration of 0.800 m/s2, find the tension in the rope (in N). N

Answer :

valeriagonzalez0213

Answer:

T= 855 N

Explanation:

Known data

m= 150 kg  : mass of the sofa

β = 30.0° : angle of the  ramp with respect to the horizontal direction

a=  0.800 m/s² : acceleration of the sofa

g = 9.8 m/s² : acceleration due to gravity

Newton's second law to the sofa:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

We define the x-axis in the direction parallel to the movement of the sofa on the ramp and the y-axis in the direction perpendicular to it.

Formula of the weight

W= m*g

x-y weight components

Wx= Wsinβ= m*g*sin30° = 150*9.8*sin30° = 735 N

Wy= Wcosβ =m*g*cos30° = 150*9.8*cos30° = 1273.05 N

Problem development

We apply the formula (1) to calculated the tension in the rope (T), and we take x + in the direction of the sofa movement:

∑Fx = m*ax  ,  ax= a  : acceleration of the sofa

T -Wx= m*a    T: tension in the rope

T= Wx+ m*a

T = 735+ 150*0.8

T= 855 N

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