4.85
Find the tension in the rope
and the acceleration of the
10 kg mass (massless rope,
frictionless pulley). No
friction between block and
table.
a
T
T
T
Force to the right on 10 kg mass:
T = ma = 10a
(1)
(2nd law)
a/2
Downward force on 3 kg mass:
3g − 2T = 3(a/2)
(2)
(2nd law)
So, from (1) and (2): 2T = 20a = 3g − 3a/2
Therefore, a = 1.37 m/s2 and T = 10a = 13.7 N
mg = 3g
Find the maximum acceleration
the truck can have without the
crate sliding off the back.
a
µs = 0.35
Fsmax = µsFN
What is FN ?
The truck and the crate are accelerated at a rate a up the slope, at
10o to the horizontal.
Rotate the axes so x is up the slope, y perpendicular to the road.
y direction (perpendicular to road)
Forces acting on the crate
FN = mg cos 10◦
So fsmax = µsmg cos 10◦
(1)
x direction (up the slope)
fsmax − mg sin 10◦ = ma (2)
Combine (1) and (2)
fsmax = µsmg cos 10◦ = ma + mg sin 10◦
a = g(µs cos 10◦ − sin 10◦)
= 9.8 × (0.35 cos 10◦ − sin 10◦)
a = 1.68 m/s2
a
Block 1: T −W1 sin 30◦ = m1a
Block 2: W2 − T = m2a
So W2 −W1 sin 30◦ = (m1 + m2)a
W2 −W1 sin 30◦
→ a=
m1 + m2
Find T by substituting for a into either of the above equations
All of the physics so far –
•
•
Newton’s 3 laws of motion
Newton’s law of gravitation
The rest is –
• useful equations – the four famous equations
• how to apply all of the above