SPIDERMAN’S

WEB FORMULA

Rules of This Activity#1 All students must perform all of the work of the

activity in their science journals, even the work that you also did on the large white board!

#2 All 4th row students will begin the activity on a large whiteboard which will then be passed to the student seated in front of them when their assigned work is completed.

#3 All previous work is to remain on the white board.

#4 The board will continue to passed to the student in the next row as each student completes their assigned work. The column of students to correctly complete the work first will be declared the winners.

The Problem

Spiderman’s new web

formula can only support

700N.

Spidey, who is 75 kg., is

suspended from two

webs , one at an angle of 30* with a force of

400N (called T1) and the

other, known as T2, is at an angle of 57*

He is only safe if the tension in

each web is less than 700N.

What is the tension in T2? Is Spiderman

safe?

Working the ProblemRow #4 Draw a force diagram of this

picture of Spiderman suspended from two webs on the whiteboard. Remember to use the four steps (C of M, long term forces, contact forces, labeled axes) in creating a force diagram!  

T2 T 1 = 400N

   

57* 30*

75 kg.

Working the ProblemRow #4 Draw a force diagram of this picture of

Spiderman suspended from two webs.

Row #3 Use SOH-CAH-TOA, to find the magnitude of the T1x , T1y, T2x , T2y and Fg

components. Show all of your work.

Working the Problem

Row #3 Use SOH-CAH-TOA, to find the magnitude of the T1x , T1y, T2x , T2y and Fg components. Show all of your work.

Row #2 Draw a “clean” force diagram, breaking Fg, T1 and T2 into their x and y components. Again remember to use the four steps (C of M, long term forces, contact forces, labeled axes) in creating a force diagram!

Working the Problem

Row #2 Draw a “clean” force diagram, breaking Fg, T1 and T2 into their x and y components.

Row #1 Calculate the tension in T2 showing all of your work. When your work is completed, consult with all of the members of your column to decide if the final answer is thought to be correct before submitting it for approval.

The Force Diagram

y T2

T1 400N

57* 30*

x

Fg

SOH – CAH - TOA

T1 = T1x + T1y

COS 30* = T1x/400N (400N) (COS 30*) = T1x

346.4N = T1x

SIN 30* = T1y/400N (400N) (SIN 30*) = T1y

200N = T1y

The “Clean” Force Diagram

y

T2y

x

T1y 200N T2x T1x 346.4N

Fg 735N

F = ma

If you used the “x” direction εFx = ma T1x - T2x = (75 kg)(0m/s2) T1x - T2x = 0 T1x = T2x

346.4N = (T2)(cos 57*) 346.4N/(cos 57*) = T2

636N = T2

or

F = ma

If you used the “y” direction εFy = ma T1y + T2y - Fg = (75 kg)(0m/s2) T1y + T2y - Fg = 0200N + (T2) (sin 57*) - (75 kg)(9.8m/s2) = 0200N + (T2) (sin 57*) – 735N = 0 (T2) (sin 57*) = 735N- 200N (T2) (sin 57*) = 535N T2 = 535N/(sin 57*) T2 = 637N

SPIDERMAN IS SAFE!