Answer:
T= 38.38 N
Explanation:
Here
mass of can = m = 3 kg
g= 9.8 m/sec2
angle θ = 40°
From figure we see the vertical and horizontal component of tension force T
If the can is to slip - then horizontal component of tension force should become equal to force of friction.
First we find force of friction
Fs= μ R
where
μ = 0.76
R = weight of can = mg = 3 × 9.8 = 29.4 N
Now horizontal component of tension
Tx= T cos 40 = T× 0.7660 N
==>T× 0.7660 = 29.4
==> T= 38.38 N
Answer:
619.8 N
Explanation:
The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:
where
T is the tension
m is the mass of the rock
v is the speed
r is the radius of the circular path
At the beginning,
T = 50.4 N
v = 21.1 m/s
r = 2.51 m
So we can use the equation to find the mass of the rock:
Later, the radius of the string is decreased to
r' = 1.22 m
While the speed is increased to
v' = 51.6 m/s
Substituting these new data into the equation, we find the tension at which the string breaks: