Tuesday, March 23rd, 2010
TA: Tomoyuki Nakayama
PHY 2048: Physic 1, Discussion Section 6839
Quiz 7 (Homework Set # 10)
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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, block A (mass 8.00 kg) is in equilibrium, but it would slip if block B (mass
6.00 kg) were any heavier. Angle θ is 25.0º.
a) Draw all the forces exerted on the block A, block B and joint of
the three cords, and write down the translational equilibrium
condition or conditions (balance of forces equations) for each of
them.
As shown in the figure, four forces are exerted on block A, two
forces are exerted on block B and three forces are exerted on the
joint. Applying translational equilibrium conditions to block A, we
get
x:TA – f = 0
y: N – mAg = 0
For block B, the translational equilibrium condition yields
TB – mBg = 0
For the joint we have
x: Tsinθ – TA = 0 y: Tcosθ – TB = 0
b) What are the tension in the cord attached to block B and that attached to block A?
Solving block B’s balance of forces equation for TB , we get
TB = mBg = 58.8 N
Solving the two balance of forces equation for the joint simultaneously, we obtain
T = TB / cosθ
TA = Tsinθ = (TB / cosθ)sinθ = TBtanθ = 27.4 N
c) What is the coefficient of static friction between block A and the surface below it?
We solve the balance of forces equation for block A to obtain the normal force and the maximum
static friction.
N = mAg = 78.4 N
f = TA = 27.4 N
By definition, the coefficient of static friction is the ratio of the maximum static friction to the normal
force, thus we have
μs = f/N = 0.349