Problem#1
The systems shown in Figure 1 are in equilibrium. If the spring scales are calibrated in newtons, what do they read? (Neglect the masses of the pulleys and strings, and assume the incline in part (c) is frictionless.) |
| Fig.1 |
(a) Isolate either mass
T + mg = ma = 0
T = |mg|
The scale reads the tension T,
So
T = mg = 5.00 kg (9.80 m/s2) = 49.0 N
(b) Isolate the pulley
T2 + 2T1 = 0
T2 = 2|T1| = 2mg = 2 x mg = 98.0 N
 |
| Fig.2 |
(c) ΣF = n + T + mg = 0
Take the component along the incline
nx + Tx + mgx = 0
or
0 + T – mg sin 30.00 = 0
T = mg sing 30.00 = 5.00 kg x 9.80 m/s2 x ½ = 24.5 N
Problem#2
Draw a free-body diagram of a block which slides down a frictionless plane having an inclination of θ = 15.0° (Fig. 3). The block starts from rest at the top and the length of the incline is 2.00 m. Find (a) the acceleration of the block and (b) its speed when it reaches the bottom of the incline.
 |
| Fig.3 |
Answer:
The two forces acting on the block are the normal force, n, and the weight, mg. If the block is considered to be a point mass and the x-axis is chosen to be parallel to the plane, then the free body diagram will be as shown in the figure to the right. The angle θ is the angle of inclination of the plane. Applying Newton’s second law for the accelerating system (and taking the direction up the plane as the positive x direction) we have
 |
| Fig.4 |
ΣFy = n − mg cos θ = 0:
n = mg cos θ
Resultant force on the x axis
ΣFx = − mg sin θ = ma
a = −g sin θ
(a) When θ =15.0°
a = −g sin θ = − 9.80 m/s2 x sin 15.00 = –2.54 m/s2
(b) Starting from rest
we use
vf2 = vi2 + 2a(xf – xi)
vf = √[2 x (–2.54 m/s2)(–2.00 m)] = 3.18 m/s
Problem#3
A 1.00-kg object is observed to have an acceleration of 10.0 m/s2 in a direction 30.0° north of east (Fig. 5). The force F2 acting on the object has a magnitude of 5.00 N and is directed north. Determine the magnitude and direction of the force F1 acting on the object.
 |
| Fig.5 |
Answer:
The acceleration experienced by an object must be parallel to the resultant force acting on the object. Therefore the direction of the resultant force acting on objects is 30.00. Then,
tan 300 = F2/F1
or
F1 = F2/tan 30.00 = 8.66 N (East)
Program#4
A 5.00-kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00-kg object, as in Figure 6. Draw free-body diagrams of both objects. Find the acceleration of the two objects and the tension in the string.
 |
| Fig.6 |
Answer:
First, consider the block moving along the horizontal. The only
force in the direction of movement is T. Thus, ∑F = max.
T = (5 kg)a
Next consider the block that moves vertically. The forces on it are
the tension T and its weight, 88.2 N.
 |
| Fig.7 |
88.2 – T = (9.0 kg)a (2)
Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be
added to give 88.2 – T = (14 kg)a. Then
a = 6.30 m/s2 and T = 31.5 N
Problem#25
A block is given an initial velocity of 5.00 m/s up a frictionless 20.0° incline (Fig. 3). How far up the incline does the block slide before coming to rest?
Answer:
 |
| Fig.8 |
∑F = max
–mg sin 20.00 = ma
a = –g sin 20.00
Taking vf = 0, vi = 5.00 m/s gives
vf2 = vi2 + 2a(xf – xi)
0 = (5.00 m/s)2 + 2(–g sin 20.00)(xf – 0)
Or
xf = 3.73 m
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