Consider two masses (M=50 kg, m = 10 kg) connected by a string as shown below. T
ID: 1900310 • Letter: C
Question
Consider two masses (M=50 kg, m = 10 kg) connected by a string as shown below. The coefficient of kinetic friction between both objects and the horizontal and inclined surfaces is ?k=0.2. Start by drawing the force diagram. g=9.8 m/s2. a) (5 pts) Find the magnitude of frictional force acting on the smaller mass b) (5 pts) Find the magnitude of frictional force acting on the larger mass c) (5 pts) Acceleration of the system d) (5 pts) Find the tension in the connecting string. e) (5 pts) If initially at rest, how long would it take for the blocks to travel 1 m?
Explanation / Answer
Consider two masses (M=50 kg, m = 10 kg) connected by a string as shown below. The coefficient of kinetic friction between both objects and the horizontal and inclined surfaces is k=0.2. Start by drawing the force diagram. g=9.8 m/s2.
a) (5 pts) Find the magnitude of frictional force acting on the smaller mass
N1 = mg = 10 * 9.8 = 98 N
fk1 = N1 = 0.2 * 98 = 19.6 N
b) (5 pts) Find the magnitude of frictional force acting on the larger mass
N2 = Mgcos30 = 50 * 9.8*cos30 = 424.35 N
fk2 = N1 = 0.2 * 424.35 = 84.87 N
c) (5 pts) Acceleration of the system
a = F/m
a = (Mgsin30- fk2 - fk1)/(M+m)
a = (50*9.8*0.5-84.87-19.6)/(50+10)
a = 2.34 m/s2
d) (5 pts) Find the tension in the connecting string.
for m:
T - fk1 = m a
>>>> T = fk1 + ma = 19.6 + 10*2.342 = 43.02 N
e) (5 pts) If initially at rest, how long would it take for the blocks to travel 1 m?
for m:
d = 0.5 a t^2
>>>> t = (2d/a) = (2*1/2.342) = 0.924 s