September 19, 2007
PHY2053
Quiz 3 (Section 4.1-4.5)
Name:
UFID:
1. (5 pts) Two people are holding a mass of 20kg with two strings attached to it. One
person pulls one of the strings with a force of 120N at an angle of 50º to the horizontal.
Find the direction of the force the other person exerts on the other string.
Express three forces in components (we take right as +x and up as +y):
Fg = (0, -20×9.8) = (0, -196)
T1 = (120cos50º, 120sin50º) = (77.1, 91.9)
T2 = (Tx, Ty)
Since Fg+T1+T2=0, we get
0+77.1+Tx = 0, Tx = -77.1
-196+91.9+Ty = 0, Ty = 104.1
θ = tanˉ¹ (104.1/-77.1)+180 = 126.5°
Opposite side to the first person and 53.5° to the horizontal.
2. (5 pts) Two crates of masses 20kg and 30kg are connected by a light string that passes
over a frictionless pulley. The 20kg crate lies on a smooth incline of angle 60º and the
30kg crate lies on a smooth incline of angle 45º. Find the tension in the string.
Applying Newton’s second law to each crate, we get
20×a1 = T - 20×9.8×sin60º = T – 170 (uphill as +x)
30×a2 = 30×9.8×sin45º-T = 208-T (downhill as +x)
Considering our choice of coordinate systems,
a1 = a2 = a (Acceleration of the whole system)
Adding the equations of motion for each crate, you get the equation of motion for the
whole system
50×a = 208-170 = 38, a = 0.76 m/s²
Substituting this in the first equation, we get
T = 170 + 20×0.76 = 185 N
3. (5 pts) A train travels horizontally with a constant acceleration. A 3.0kg mass hangs
at one end of a string that is attached to the ceiling of the train. A passenger has to push
the object horizontally with the force of 8.0N to keep the string vertical. What is the
acceleration of the train?
Using F = ma, we get
a = F/m = 8/3 = 2.67 m/s²
Note that tension and gravity don’t have horizontal components.
4. (5 pts) A roller coaster travels down an inclined track of an angle of 60º with an
constant acceleration. A passenger on the coaster holds an end of a string with a 0.5kg
mass attached to the other end. The string makes an angle of 15º to the vertical. What is
the acceleration of the coaster?
Since the gravity and tension are exerted on the mass, we use Fg+T = ma. Express this
equation in components (take +x downhill and +y perpendicular to the incline):
0.5×9.8×sin60°-Tcos45° = 0.5a
-0.5×9.8cos60º+Tsin45° = 0, T = 4.9cos60°/sin45°
Plugging T into the first equation, you get
0.5a = 4.9sin60°-4.9cos60°/tan45°, a = 3.59 m/s²