Problem 13

A bird feeder is connected as shown. Find the tension in each cable.

SOLUTION:

There will be two free-body diagrams. The first on is for the bird feeder and the second one is for the knot where all three cables are attached. The diagrams are shown in the figures below.

Bird feeder equations:

            SFy = 0            the system is in equilibrium

            T3 – mg = 0

            T3 = mg = 150 N                                 (equation 1)

The knot equations:

            SFy = 0

            T1 sin(600) + T2 sin(300) – T3 = 0

            T1 sin(600) + T2 sin(300) = 150 N         (equation 2)

            SFx = 0

            - T1 cos(600) + T2 cos(300) = 0

            T1 = (1.73) T2                                                   (equation 3)

Now we just solve the equations for T1 and T2 :

            From (1)          (0.866)T1 + (0.500)T2 = 150 N

                                    (0.866)(1.73)T2 + (0.500)T2 = 150 N

                                    1.498 T2 + 0.500 T2 = 150 N

                                    2.0 T2 = 150 N

                                    T2 = 75 N

                                    T1 = 130 N

 

Problem 16

Two persons are pulling a boat through water as in the figure below. Each exerts 600 N directed at 300 angle relative to the forward motion of the boat.If the boat moves with a constant velocity, find the resistive force F.

SOLUTION:

Draw the free-body diagram and realize that the system is in equilibrium since the system is moving with constant velocity.

            SFx = 0

            (600 N)cos(300) + (600 N)cos(300) – F = 0

            F = 2(600 N)cos(300)

            F = 1039 N

Problem 17

A 5.0-kg bucket of water is raised from a well by a rope. If the upward acceleration of the bucket is 3.0 m/s2, find the force exerted by the rope on the bucket.

SOLUTION:

            SFx = 0

            T – mg = ma

            T = mg + ma

            T = m(g + a) = (5.0 kg) (9.80 m/s2 + 3.0 m/s2) = 64 N

 

Problem 26

Two blocks are fastened to the ceiling of an elevator as in the figure. The elevator accelerates upward at 2.00 m/s2.Find the tension in each rope.

SOLUTION:

The free-body diagrams are shown in the figures. There are 2 pieces of rope therefore there are 2 different tensions we need to calculate.

Object 1:

            SFy = ma

            T1 – mg – T2 = ma

Object 2:

            SFy = ma

            T2 – mg = ma

The next step is algebra.

            T2 = (10 kg)(m + a) = (10 kg)(9.80 + 2.00)m/s2 = 118 N

            T1 = mg + ma + T2 = 118 N + 118 N = 236 N

 

Problem 28

Two masses of 3.00 kg and 5.00 kg are attached as shown in the figure. Determine (a) the tension in the string, (b) the acceleration of each mass, (c) the distance each mass will move in the first second of motion if both masses start from rest.

SOLUTION:

Draw the free-body diagram with the accelerations directed up for the smaller mass and down for the larger mass.

3.00 kg Mass:

                        SFy = ma

                        T – mg = ma

                        T = (3.00 kg)a + (3.00 kg)g

5.00 kg Mass:

                        SFy = -ma

                        T – mg = -ma

                        T = mg – ma

                        T = (5.00 kg)g – (5.00 kg)a

Solve the equations simultaneously:

                        (3.00 kg)a + (3.00 kg)g = (5.00 kg)g – (5.00 kg)a

                        (8.00 kg)a = (2.00 kg) g

                        a = 0.25 g = 2.45 m/s2

 

Problem 36

A student decides to move a box of books into her dorm room by pulling on a rope attached to the box. She pulls with a force of 80.0 N at an angle of 25.00 above the horizontal. The box has a mass of 25.0 kg., and the coefficient of kinetic friction between the box and the floor is 0.300. (a) Find the acceleration of the box. (b) The student now starts moving up a 10.00 incline, keeping her 80.0 N force steady and the angle still 25.00 above the line of the incline. If the coefficient of friction is now unchanged, what is her new acceleration?

SOLUTION:

(a) The system is in equilibrium in the y-direction but not in the x-direction.

            SFx = ma

            80.0 N cos 25.00 – f = ma

            72.5 N – f = ma

            SFy = 0

            FN + 80.0 sin 25.00 – mg = 0

            FN = 211.19 N

            f = mFN = (0.300)(278.81 N)

            f = 63.36 N

            a = (1/25.0 kg) (72.5 N – 63.36 N) = 0.366 m/s2

(b) The system is still in equilibrium in the y but not in the x-direction.

            SFx = ma

            80.0 N cos 25.00 – f– mg sin 10.00= ma

            72.5 N – 42.5 N – f = (25.0 kg)a

            30.0 N – f = (25.0 kg)a

            SFy = 0

            FN + 80.0 sin 25.00 – mg cos 10.00 = 0

            FN = 207.5 N

            f = mFN = (0.300)(207.5 N)

            f = 62.2 N

            a = -1.29 m/s2

 

Problem 41

Find the acceleration of each of the two masses shown in the figure if the coefficient of friction between the 7.00-kg mass and the plane is 0.250.

SOLUTION:

            SFx = ma

            T – f– mg sin q = ma

            SFy = 0

            FN – mg cos q = 0

            FN = mg cos q

            f = mFN = mmg cos q

HANGING MASS

            T – mg = -ma

            T = mg – ma = 117.6 N – (12.0 kg)a

Algebra:

            T – f– mg sin q = ma

117.6 N – (12.0 kg)a – mmg cos q– mg sin q = ma

117.6 N – 12.0 a – 13.7 N – 41.3 N = (7.00)a

19.0 a = 62.6 N

a = 3.30 m/s2