A scaffold of mass 68 kg and length 7.0 m is supported in a horizontal position

Question

A scaffold of mass 68 kg and length 7.0 m is supported in a horizontal position by a vertical cable at each end. A window washer of mass 79 kg stands at a point 1.8 m from one end. What is the tension in the cable closer to the painter? What is the tension in the cable further from the painter? Did you pick a pivot point about which to calculate torques? Do you see that with an unknown force at each end, that one of the ends should be the pivot point? Did you write the torques about the pivot for those two unknown forces and for the force on the scaffold due to the window washer's weight? Did you include the correct signs? After you find one of the unknown forces, did you write A balance-of-forces equation for the vertical axis?

Explanation / Answer

Let T1 and T2 be tensions in the two cables

Weight of scaffold

W1=68*9.8 =666.4 N

Weight of man

W2=79*9.8=774.2 N

At equilibrium net torque is zero

7T2-(7/2)*666.4-1.8*(774.2)=0

T2=532.3 N

Y-component

T1+T2-666.4-774.2=0

T1=908.3 N

a)

T1=908.3 N

b)

T2=532.3 N

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