Help with physics newtons law?

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1 Answer
Mar 19, 2018

2a) see below

2b) #F_(T1) = 919.7 "N"# at #155^o# and #F_(T2) = 846.1 "N"# at #10^o#

Explanation:

2a) A free body diagram shows all forces acting on an object, in this case the climber. There are 3 forces acting on the climber: the rope tension on the left, the rope tension on the right, and the force of gravity acting downwards (note: since the magnitude was told to be ignored it is not accurate for that)

The angles are given as a counterclockwise rotation from the #0^o# direction.

2b) The tensions in the rope have to counteract each other horizontally and also counteract the force of gravity vertically. Each of these problems should be handled separately as forces acting in perpendicular directions are independent.

We can draw two right triangles to give us a sense of geometry for this problem.

Let's first compute #F_g# which is the weight of the climber:

#=> F_g = mg = (54.6)(9.81) = 535.6 " N"#

The other forces we care about are:

  • # F_(T1_x) " - the horizontal tension of rope segment on left"#
  • # F_(T1_y) " - the vertical tension of rope segment on left"#
  • # F_(T2_x) " - the horizontal tension of rope segment on right"#
  • # F_(T2_y) " - the vertical tension of rope segment on right"#

Since the state of the system is static, there is no net force acting on the climber when she is resting. So all forces in the horizontal and vertical must cancel.

Vertical (let up be positive and down be negative)
#F_(T1_y) + F_(T2_y) - F_g = 0#

#F_(T1)sin(90^o-65^o)+F_(T2)sin(90^o-80^o)-535.6=0#

#=>F_(T1)sin(25^o)+F_(T2)sin(10^o)= 535.6#

Horizontal (let right be positive and left be negative)
#-F_(T1_x) + F_(T2_x) = 0#

#-F_(T1)cos(25^o) + F_(T2)cos(10^o) = 0#

#F_(T2)cos(10^o) = F_(T1)cos(25^o)#

#F_(T2) = cos(25^o)/cos(10^o) F_(T1) #

#=>F_(T2)= 0.92 F_(T1)#

We can now use this result in the vertical equation:

#F_(T1)sin(25^o)+F_(T2)sin(10^o)= 535.6#

#F_(T1)sin(25^o)+(0.92 F_(T1))sin(10^o)= 535.6#

#F_(T1)[sin(25^o) +0.92sin(10^o)] = 535.6#

#F_(T1) = 535.6/(sin(25^o) +0.92sin(10^o))#

#=> color(blue)(F_(T1) = 919.7 " N")#

Now we can go back and find the second force:

#F_(T2) = 0.92 F_(T1)#

#F_(T2) = 0.92 (919.7)#

#=> color(blue)(F_(T2) = 846.1 " N")#

Adding in the angles from the free body diagram we have:

#=>color(green)(F_(T1) = 919.7 "N" " at " 155^o)#

#=>color(green)(F_(T2) = 846.1 "N" " at " 10^o)#