TA: Tomoyuki Nakayama Monday, March 22nd, 2010
PHY 2048: Physic 1, Discussion Section 3081
Quiz 7 (Homework Set # 10)
Name: UFID:
Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit.
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In the figure below right, horizontal scaffold 2, with uniform mass m2 = 20.0 kg and length L2 = 2.00 m, hangs from horizontal scaffold 1, with uniform mass m1 = 40.0 kg. A 30.0 kg box of nails lies on scaffold 2, centered at distance d = 0.500 m from the left end.
a) Draw all the forces exerted on the lower scaffold in the figure, and write down the translational equilibrium condition (balance of forces equation) and rotational equilibrium condition (balance of torques equation) for the lower scaffold.
Four forces are exerted on the lower scaffold, the gravitational force exerted at the center of the scaffold, two tension forces exerted at the both ends and the force from the box, whose magnitude is equal to the gravitational force on the box.
The translational equilibrium condition yields
TL + TR – m2g – Mg = 0
Choosing our rotational axis at the left end of the lower scaffold, the rotational equilibrium condition yields
-Mgd - m2g(L2/2) + TRL2 = 0
b) Solve the equations you set up in part a) and find the tensions in the left cable and right cable supporting the lower scaffold.
Solving the rotational equilibrium condition for TR, we get
TR = [Mgd + m2g(L2/2)]/ L2 = 171.5 N
Plugging the value of TR into the translational equilibrium condition, we get
TL = -TR + m2g + Mg = 318.5 N
c) What is the tension in the right cable supporting the upper scaffold?
We repeat the procedures in part a) and b) for scaffold 2. Five forces are exerted on the upper scaffold as shown in the FBD on the right. Choosing our rotational axis at the left end of the scaffold, the rotational equilibrium condition yields
-TLd-m1g(d+L2/2)-TR(d+L2) +T’R(L+2d) = 0
⇒ T’R = [TLd + m1g(d + L2/2) +TR(d + L2)] / (L+2d) = 392 N