Physics lesson, tension force on ropes. Questions and solutions related to tension force in ropes. To be found of tension force of ropes on inclined plane.

Question 1 


System in the figure is balanced with T1 , T2, T3 ropes and K object.

Tension force on the T2 rope is 60 N. 

Find the mass of the "K" object.

Find the tensile force of the T1 and T2 ropes.

(g = 10 m/s2, sin37° = 0.6, cos 37°= 0.8)

Solution 1 


The tension force of the T2 rope is equal, to the component of on x axis of the tension force in T1 rope.

T2 = T1x

As in the figure the T1x vector is one of perpendicular edge of the right triangle. So it is the component of on the x axis of T1 vector.

cos37° = T1x

0.8= 60

T1 = 75 N

Tension force of the T3 rope.

We create a right triangle with T1 and T3 ropes. The hypotenuse of this triangle will be T3 rope.

sin37° = T1

0.6 = 75

T3 = 125 N.

The weight of the "K" object is equal to the tension force in the T3 rope.

WK = m.g

125 = m •10

m = 12,5 kg

Question -2


The inclined plane in the figure is frictionless.

The mass of the K body is 5 kg, and the mass of the L body is 8 kg.

A) Find the acceleration of the system.

B) Find the tension forces at the ropes T1 and T2 .

Solution 2 


The T1 and the T2 ropes are the same rope. It’s tension forces are equal.

The FL is gravity force that effect to L object.

The FL force is transmitted exactly to the K object by the rope.

There are two forces that effect to K object. One of it’s forces is the Fx force that pulls the K object to downward.

The other force is the FL force which pulls upward the K object parallel to the inclined plane.

The K object moves in direction of component of the forces Fx and FL.

FL = 8 •10 = 80 N

The F force is equal to the weight of the K object and it is in the direction of gravity.

F = WL 

WL = m• g

WL = 5 • 10 = 50 N

The Fy force is the perpendicular component of the gravity force on inclined plane. It is not considered at calculates, because the system is frictionless.

The Fx force is a parallel component of the gravitational to inclined plane. It is be found as below.

cos53 = Fx

0.6 = Fx

Fx = 30 N


A) The acceleration of the system.

Fnet = m•a

FL - Fx = m•a

80 - 30 = (8 + 5) • a

50 = 13•a

a = 3.85 m/s2

B) Tension of the ropes T1 and T2.

On the "K" object.

Fnet = m•a

T2 –Fx = 5 • 3.85

T2 –30 = 19,25

T2 = 49,25 N

Fx and FL forces pull the T2 rope to opposite directions. Therefore, the tension force at the T2 rope is the sum of Fx and FL.

The tension force of the T1 rope is equal the tensile force of the rope T2.

T1 = 49,25 N

Question 3 


Seen in the figüre the K rod is could rotate at around the O point. The K rod is weightless and its segments are equal length.

P1 and P2 objects are hanging as in the figüre.

The system is in balance.

Find the magnitude of the tensile force on the T rope.

(sin 37° = 0,6)

Solution 3 

We Show the forces on the rod as follows,


The P1 and the P2 masses pull downward the rod, the Ty force pulls to upward the rod. Since the forces pulling the rod down are equal to the forces pulling up, the system is balanced.

We could write the equation below,

2•P1 + 3•P2 = 4•Ty

2•15 + 3•10 = 4 • Ty

60 = 4•Ty

Ty = 15 N

Let we find T1 rope tensile now.

The Ty is vertical component of T1 vector and it is found as below,

Ty = sin37°

Ty = T.0.6

T = 15

T = 25 N

Tension Force Questions In Rope. The Part 1

Tension Force Questions In Rope. The  Part 3



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