March 28, 2008
PHY2053 Discussion
Quiz 8 (Chapter 8.1-8.5)
Name:
UFID:
*1. (5 pts) A 4.00 ft by 8.00 ft uniform rectangular sheet of plywood is located so that xand y- coordinates of the four vertices are (4.00, 2.00) ft, (-4.00, 2.00) ft, (-4.00, -2.00) ft
and (4.00, -2.00) ft. Now suppose the upper right quadrant is removed from the sheet
and find the center of the gravity of the remaining sheet.
We consider the sheet consisting of two parts: One is a 4 ft by 4 ft square in the 2nd and
3rd quadrants, the other one is a 4 ft by 2 ft rectangle in the 4th quadrant. Their centers
of gravity are located at (-2, 0) ft and (2, -1) ft respectively. The 4×4 square is twice as
heavy as 4×2 rectangle. Thus the center of the gravity of the remaining sheet is located
at
X = (2m×(-2)+m×2)/(2m+m) = -2/3 = -0.667 ft
Y = (2m×0+m×(-1))/(2m+m) = -1/3 = -0.333 ft
**2. (5 pts) A 5.00 m, 20.0 kg uniform ladder rests against a frictionless wall. The ladder
makes 60.0º to the horizontal ground. A 60.0 kg person climbs up the ladder, and when
he/she moves up 3.00 m, the ladder slips. What is the coefficient of static friction
between the base of the ladder and the ground?
Since the ladder is on the verge of slipping, friction equals the coefficient of static
friction times normal force. Applying Newton’s second law, we have
X: f-n = 0 ⇒ μN-n = 0 ⇒ μ = n/N
Y: N-mg-Mg = 0 ⇒ N = (m+M)g = (60+20)×9.8 = 784 N
Net torque around the base of the ladder is zero, thus we have
nLsinθ-mglcosθ-Mg(L/2)cosθ = 0
⇒ n = (ml+M(L/2))gcosθ)/ Lsinθ = (60×3+20×2.5)9.8×cos60˚/5×sin60˚ = 260 N
Thus, the coefficient of static friction is given by
μ = n/N = 260/784 = 0.332
*3. (5 pts) A baton consists of a 0.8 kg solid rod and two 0.4 kg masses attached to the
ends of the rod. The rod is 1.00 m long. Determine the moment of inertia of the baton
about an axis through its center of the rod and perpendicular to its length. The moment
of inertia of a solid rod around the center is given by (1/12)ML².
The moment of inertia of the baton is the sum of the moment of inertia of the rod and
that of two masses. Thus we have
I = (1/12)ML²+2m(L/2)² = (1/12)×0.81+2×0.4×0.5² = 0.267 kgm²
***4. (5 pts) A 4.00 kg block on a 30.0º incline is attached to a light string that passes
over a 2.00 kg pulley at the top of the incline and then connected to an 8.00 kg hanging
object. The coefficient of kinetic friction between the surface and the block is 0.200. The
pulley is a solid cylinder and has a radius of 0.200 m. The string does not slip on the
pulley. What is the acceleration of the block? (The moment of inertia of a solid cylinder
is (1/2)MR².)
Since the string does not slip on the pulley, the angular acceleration of the pulley is
related to the linear acceleration of the block by α = a/R. Taking positive direction to
the direction of motion, the equations of motion of the block, the hanging object and the
pulley are given by
ma = T-mgsinθ-μmgcosθ
m’a = m’g-T’
Iα = (T’-T)R ⇒(1/2)MR²(a/R) = (T’-T)R ⇒ (1/2)Ma = T’-T
Add all three equations and solve for the acceleration:
(m+m’+(1/2)M) = m’g- mgsinθ-μmgcosθ ⇒ a = (m’-m(sinθ+μcosθ))g/( m+m’+(1/2)M)
= (8-4(sin30˚+0.2×cos30˚)×9.8/(4+8+0.5×2) = 4.00 m/s²