Consider the two-body system at the right. A 22.7-N block is placed upon an inclined plane which is inclined at a 17.2 degree an
gle. The block is attached by a string to a 34.5-N block which is suspended over a pulley. The coefficient of friction is 0.219. Determine the acceleration of the block and the tension in the stringAnswers:
1 answer:
Answer:
Given that
For A weight Wt= 22.7 N
m₁ = 22.7/10 = 2.27 kg
Force alone inclined plane
Wt₁ = m₁ g sin θ
Wt₁ =22.7 sin 17.2°
Wt₁ = 6.7 N
For B weight Wt₂= 34.5-N
m₂ = 3.45 kg
coefficient of friction ,μ= 0.219
θ = 17.2 degree
The friction force on the block A
Fr= μ m₁ g cos θ
Fr= 0.216 x 22.7 x cos 17.2°
Fr= 4.68 N
Lets take acceleration of system is a m/s²
Tension = T
From Newtons law
Wt₂ - Wt₁ - Fr = (m₁ +m₂) a
34.5 - 6.7 - 4.68 = (2.27 + 3.45 ) a
a= 4.05 m/s²
Block B
Wt₂ - T = m₂ a
T = 34.5 - 4.05 x 3.45
T= 20.52 N
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On further solving
1) In a circular motion, the angular displacement 
is given by
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire:
. Using the information about the radius, 
, we find the total angular displacement:
2) If we put larger tires, with radius
, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire:
2) If we put larger tires, with radius