Download Block on an Incline Adjacent to a Wall

skiladæmi 3
Due: 11:59pm on Wednesday, September 23, 2015
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Block on an Incline Adjacent to a Wall
A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane, as shown. There is
no friction between the wedge and the block or between the wedge and the horizontal surface.
Part A
Find the magnitude, Fnet , of the sum of all forces acting on the block.
Express Fnet in terms of θ and m, along with any necessary constants.
Hint 1. Direction of the net force on the block
The net force on the block must be the force in the direction of motion, which is down the incline.
Hint 2. Determine the forces acting on the block
What forces act on the block? Keep in mind that there is no friction between the block and the wedge.
ANSWER:
The weight of the block and friction
The weight of the block and the normal (contact) force
The weight of the block and the weight of the wedge
The weight of the block and the force that the wall exerts on the wedge
Hint 3. Find the magnitude of the force acting along the direction of motion
Consider a coordinate system with the x direction pointing down the incline and the y direction perpendicular to
the incline. In these coordinates, what is wx , the component of the block's weight in the x direction?
Express wx in terms of m, g, and θ.
ANSWER:
wx
= mgsin(θ)
ANSWER:
F net
= mgsin(θ)
Correct
Part B
Find the magnitude, Fww , of the force that the wall exerts on the wedge.
Express Fww in terms of θ and m, along with any necessary constants.
Hint 1. The force between the wall and the wedge
There is no friction between the wedge and the horizontal surface, so for the wedge to remain stationary, the net
horizontal force on the wedge must be zero. If the block exerts a force with a horizontal component on the
wedge, some other horizontal force must act on the wedge so that the net force is zero.
Hint 2. Find the normal force between the block and the wedge
What is the magnitude, n, of the normal (contact) force between the block and the wedge? (You might have
computed this already while answering Part A.)
Express n in terms of m, g, and θ.
ANSWER:
= mgcos(θ)
n
Hint 3. Find the horizontal component of the normal force
In the previous hint you found the magnitude of the normal force between the block and the wedge. What is the
magnitude, nh , of the horizontal component of this normal force?
Express nh in terms of θ and n.
ANSWER:
nh
= nsin(θ)
ANSWER:
F ww
= mgsin(θ)cos(θ)
Correct
Your answer to Part B could be expressed as either mg sin(θ) cos(θ) or mg sin(2θ)/2. In either form, we see
that as θ gets very small or as θ approaches 90 degrees (π/2 radians), the contact force between the wall and the
wedge goes to zero. This is what we should expect; in the first limit ( θ small), the block is accelerating very
slowly, and all horizontal forces are small. In the second limit ( θ about 90 degrees), the block simply falls
vertically and exerts no horizontal force on the wedge.
Problem 5.73
An 3.00­kg box sits on a ramp that is inclined at 33.0∘ above the horizontal. The coefficient of kinetic friction between the
box and the surface of the ramp is μk = 0.300. A constant horizontal force F = 26.0 N is applied to the box , and the box
moves down the ramp.
Part A
If the box is initially at rest, what is its speed 2.00 s after the force is applied?
Express your answer with the appropriate units.
ANSWER:
v
= 23.1 m
s
Correct
± Mass on Turntable
A small metal cylinder rests on a circular turntable that is rotating at a constant speed as illustrated in the diagram .
The small metal cylinder has a mass of 0.20 kg, the coefficient of
static friction between the cylinder and the turntable is 0.080, and
the cylinder is located 0.15 m from the center of the turntable.
Take the magnitude of the acceleration due to gravity to be 9.81 2
m/s .
Part A
What is the maximum speed v max that the cylinder can move along its circular path without slipping off the turntable?
Express your answer numerically in meters per second to two significant figures.
Hint 1. Centripetal acceleration
If you know a body is in uniform circular motion, you know what its acceleration must be. If a body of mass m
is traveling with speed v in a circle of radius R, what is the magnitude ac of its centripetal acceleration?
ANSWER:
mv 2
R
mv 2 R v 2 R v
2
R
Hint 2. Determine the force causing acceleration
Whenever you see uniform circular motion, there is a real force that causes the associated centripetal
acceleration. In this problem, what force causes the centripetal acceleration?
ANSWER:
normal force
static friction
weight of cylinder
a force other than those above
Hint 3. Find the maximum possible friction force
The magnitude f s of the force due to static friction satisfies f s
≤ f
max
. What is f max in this problem?
Express your answer numerically in newtons to three significant figures.
ANSWER:
f
max
= 0.157 N Hint 4. Newton's 2nd law
To solve this problem, relate the answers to the previous two hints using Newton's 2nd law:
⃗ F = m a⃗ .
ANSWER:
v max
= 0.34 m/s Correct
Two Blocks and Two Pulleys
A block of mass m1 is attached to a massless, ideal string. This string wraps around a massless pulley and then wraps
around a second pulley that is attached to a block of mass m2 that is free to slide on a frictionless table. The string is
firmly anchored to a wall and the whole system is frictionless.
Use the coordinate system indicated in the figure when solving this problem.
Part A
Assuming that a2 is the magnitude of the horizontal acceleration of the block of mass m2 , what is T , the tension in
the string?
Express the tension in terms of m2 and a2 .
Hint 1. Which physical principle to use
You should use Newton's 2nd law:
→
→
→
→
F 21 + F 22 +. . . +F 2n = m 2 a2
,
⃗ ⃗ where F 21
, F 22
, ... are forces acting on the block of mass m2 . Keep in mind that the whole system is
frictionless.
Hint 2. Force diagram for the block of mass m2
Which figure correctly illustrates the forces acting on the block of mass m2 ?
⃗ ⃗ The vectors n⃗ , w⃗ , T , and f denote the normal force,
the gravitational force, the tension in the string, and
the friction force, respectively.
ANSWER:
Figure 1
Figure 2
Figure 3
None of the above
ANSWER:
T
= m 2 a2
2
Correct
Part B
Given T , the tension in the string, calculate a1 , the magnitude of the vertical acceleration of the block of mass m1 .
Express the acceleration magnitude a1 in terms of m1 , g, and T .
Hint 1. Which physical principle to use
Apply Newton's 2nd law:
→
→
→
→
F 11 + F 12 +. . . +F 1n = m 1 a1
,
⃗ ⃗ where F 11
, F 12
, ... are forces acting on the block of mass m1 .
Hint 2. Force diagram for the block of mass m1
Which figure correctly illustrates the forces acting on the block of mass m1 ?
⃗ ⃗ The vectors w⃗ , T , and F i denote the gravitational
force, the tension in the string, and the inertial force,
respectively.
ANSWER:
Figure 1
Figure 2
Figure 3
None of the above
ANSWER:
a1
= −
T
m1
+g
Correct
Part C
Given the magnitude a1 of the acceleration of the block of mass m1 , find a2 , the magnitude of the horizontal
acceleration of the block of mass m2 .
Express a2 in terms of a1 .
Hint 1. Method 1: String constraint (uses calculus)
Define y1 (t) and x2 (t) as the vertical coordinate of the block of mass m1 and the horizontal coordinate of the
block of mass m2 , respectively. It is clear that L, the
length of the string, is
L = 2x2 (t) − y (t) + C
1
,
where C is a constant that accounts for the wound
portions of the string and the length of string between
the y axis and the wall. Do not worry about the value
of C , as it will vanish in the next step. By
differentiating this equation twice with respect to time,
you should obtain a relation between a1 and a2 . The
variables C and L will vanish upon differentiation.
Hint 2. Method 2: Intuition (does not involve calculus)
You should notice that, while the block of mass m1 descends a height h, the other will move only half of h.
Hence, at each instant, v 1 = 2v 2 , where v 1 and v 2 are the speeds of the blocks of masses m1 and m2 ,
respectively. The formula for a2 versus a1 should be obvious.
ANSWER:
a2
= 1
2
a1
Correct
Part D
Using the result of Part C in the formula for T that you previously obtained in Part A, express T as a function of a1 .
Express your answer in terms of m2 and a1 .
ANSWER:
T
= m2 a1
1
4
Correct
Part E
Having solved the previous parts, you have all the pieces needed to calculate a1 , the magnitude of the acceleration of
the block of mass m1 . Write an expression for a1 .
Express the acceleration magnitude a1 in terms of m1 , m2 , and g.
Hint 1. How to approach this problem
In Part B, using Newton's 2nd law, you derived a relation between a1 and the tension in the string, T . In Part D
you found T as a function of a1 . Now eliminate T from this system of two linear equations and solve for a1 .
ANSWER:
a1
= 4gm 1
m 2 +4m 1
Correct
Score Summary:
Your score on this assignment is 102%.
You received 4.07 out of a possible total of 4 points.