Physics/Maths

Question

A seat of mass 3 kg is suspended by a rope that is attached to a tree
branch. The seat is pushed sideways, so that the rope remains taut, and
makes an angle of 55? with the downward vertical. The pushing force
makes an angle of 25? with the upward vertical. The seat is at rest

 

find the magnitude of the pushing force acting on the seat

Answer 1

First identify the external forces acting on the seat. There is the tension T due to the rope, the gravitational force Fg (of magnitude 3g), and the pushing force, call it F. Since the seat is at rest, F(net,x) = 0 and F(net,y) = 0.  We have

 

0 = F(net,x) = F sin(25) - T sin(55)  (1)

 

0 = F(net,y) = F cos(25) + T cos(55) - 3g  (2)

 

By (1),

 

F sin(25) = T sin(55)  (3)

 

By (2),

 

3g - F cos(25) = T cos(25)  (4)

 

Using (3) and (4), we write the proportion

 

F sin(25)/[3g - F cos(25)] = T sin(55)/[T cos(25)],

 

which is the same as

 

F sin(25)/[3g - F cos(25)] = tan(55) (5)

 

Multiplying (5) by 3g - F cos(25), we have

 

F sin(25) = 3g tan(55) - F cos(25)tan(55),

 

F sin(25) + F cos(25)tan(55) = 3g tan(55),

 

F[sin(25) + cos(25)tan(55)] = 3g tan(55).

 

Therefore

 

F = 3g tan(55)/[sin(25) + cos(25)tan(55)].

 

Since the pushing force makes an angle of 25 degrees with the upward vertical, the force vector is 

 

-F sin(25) i + F cos(25) j = -{3g sin(25) tan(55)/[sin(25) + cos(25)tan(55)]} i + {3g cos(25) tan(55)/[sin(25) + cos(25)tan(55)]} j.