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Dynamics 1e 237
Problem 2.
A woman is sliding down an incline with a constant acceleration of a 0 D 2:3 m=s^2 relative to the incline. At the same time the incline is accelerating to the right at 1:2 m=s^2 relative to the ground. Letting D 34 ı^ and L D 4 m and assuming that both the woman and the incline start from rest, determine the horizontal distance traveled by the woman with respect to the ground when she reaches the bottom of the slide.
Solution
The time it takes the person to slide down the incline is found using the constant acceleration equation
s D s 0 C v^20 C 12 ac t^2 ) L D 12 a 0 t^2 ) t D
s 2L a 0
Adopting a Cartesian coordinate system with O{ in the direction of as and |O opposite to the direction of gravity, we can describe the acceleration of the incline and the acceleration of the person relative to the incline as follows: a Es D as O{ and Eap=s D a 0 .cos O{ C sin | /:O (2)
Therefore, using relative kinematics, we have
a Ep D Eas C Eap=s D .as a 0 cos /O{ a 0 sin | :O (3)
The horizontal distance covered by the person is found with the same constant acceleration equation used earlier but with the acceleration set to the horizontal component of aEp. This yields
xp D 12 .as a 0 cos / t^2 ) xp D
L
a 0
.as a 0 cos / D 1:23 m:
where t was replaced by its expression given in the last of Eqs. (1) and where we have used the following numerical data: L D 4 m, a 0 D 2:3 m=s^2 , as D 1:2 m=s^2 , and D 34 ı. In summary, we can say
The person travels 1:23 m to the left of where she started.
Problem 3.
The crate A of mass m and the wedge B on which it rests are both initially at rest. The wedge, whose face is inclined at ✓ D 30 ı^ with the horizontal, is given an acceleration aB to the left as shown. Given that the coefficient of static friction between the crate and the wedge is s D 0:6, determine the maximum value of aB such that the crate does not slip on the wedge.
Solution
To determine the maximum acceleration so that the crate does not slip we begin by considering the case in which slip down the incline is impending. Referring to the FBD at the right, we model the crate as a particle subject to its own weight mg, the normal force N , and the force F due to friction. The direction of F is due to the stated impending slip condition. The latter will be to be verified by checking whether the corresponding solution confirms that aBmax > 0, F > 0, and N > 0. If not, another working assumption will be made and a new corresponding solution will be obtained.
Balance Principles. Applying Newton’s second law, we have
X FxW N sin ✓ F cos ✓ D maAx ; (1) X FyW N cos ✓ C F sin ✓ mg D maAy ; (2)
where aAx and aAy are the x and y components of the the acceleration of the crate, respectively.
Force Laws. Because of the stated impending slip condition we have
F D s N: (3)
Kinematic Equations. Since the crate does not slip, it accelerates with the incline in the horizontal direction with aB D aBmax : aAx D aBmax and aAy D 0: (4)
Computation. Substituting Eqs. (3) and (4) into Eqs. (1) and (2), we have
N sin ✓ s N cos ✓ D maBmax and N cos ✓ C s N sin ✓ mg D 0: (5)
These equations form a system of two equations in the two unknowns aBmax and N whose solution is
aBmax D
g. s cos ✓ sin ✓/ cos ✓ C s sin ✓
and N D
mg cos ✓ C s sin ✓
Recalling that g D 9:81 m=s 2 D 32:2 ft=s 2 , s D 0:6, and ✓ D 30 ı, we can evaluate the result in the first of Eqs. (6) to obtain
aBmax D 0:1650 m=s^2 D 0:5417 ft=s^2 :
Referring to the second of Eqs. (6), the term cos ✓ C s sin ✓ D 1:166 > 0 so that N > 0 and F > 0, as expected.
612 Solutions Manual
Problem 3.
A 62 kg woman A sits atop the 60 kg cart B, both of which are initially at rest. The cart is rigidly attached to a wall by the rope CD. If the woman slides down the frictionless incline of length L D 3:5 m, determine the tension in the rope CD as she slides down the incline. Ignore the mass of the wheels on which the cart can roll. The angle ✓ D 26 ı.
Solution
We model the woman A and the cart B as particles. We neglect friction between A and B, A is assumed to be subject to her own weight mAg and the normal reaction NAB from B. The incline B is assumed to be subject to its own weight mB g, the normal reaction NAB , the normal reaction N with the ground and the tension in the cord T.
Balance Principles. Applying Newton’s second law, we have ⇣X Fx
A
W NAB sin ✓ D mAaAx ; (1) ⇣X Fy
A
W NAB cos ✓ mAg D mAaAy ; (2) ⇣X Fx
B
W T NAB sin ✓ D mB aBx ; (3) ⇣X Fy
B
W N NAB cos ✓ mB g D mB aBy : (4)
Force Laws. All forces are accounted for on the FBDs.
Kinematic Equations. The inline B is stationary. Hence, letting aA denote the magnitude of the accelera- tion of A, we have
aAx D aA cos ✓; aAy D aA sin ✓; aBx D 0; aBy D 0: (5)
Computation. Substituting Eqs. (5) into Eqs. (1)–(4), we have
NAB sin ✓ D mAaA cos ✓; NAB cos ✓ mAg D mAaA sin ✓; (6) T NAB sin ✓ D 0; N NAB cos ✓ mB g D 0; (7)
which is a system of four equations in the four unknowns T , N , NAB and aA whose solution is
T D mAg sin ✓ cos ✓; N D 12 ŒmA.1 C cos^2 ✓ sin^2 ✓/ C 2mB çg; (8) NAB D mAg cos ✓; aA D g sin ✓: (9)
Observing that T is constant, and recalling that mA D 62 kg, ✓ D 26 ı, and g D 9:81 m=s^2 , we can evaluate T to obtain T D 239:6 N:
