# Physics 1351

## 25 September1997

Possibly useful information:

s = si + vi t + (1/2) a t2

v = vi + a t

v2 = vi2 + 2 a (s - si)

sin = opp/hyp

F = m a

F12 = - F21

fsmax = µs n

fk = µk n

g = 9.8 m/s2 = 10 m/s2

 1. A hot air balloon is traveling vertically upward at a constant speed of 5.0 m/s. When it is 10.0 m above the ground, a sandbag is released from the baloon. (a) After it is released, for how long is the sandbag in the air? (b) What is its velocity just before impact with the ground!

(a) After it is released, for how long is the sandbag in the air?

y = yi + vi t + (1/2) ay t2

y = yi+ vi t + (1/2) (- g) t2

0 = 10 m + (5 m/s) t + (1/2) (- 10 m/s2) t2

0 = 10 + 5 t + (1/2) (- 10) t2

0 = 10 + 5 t - 5 t2

5 t2 - 5 t - 10 = 0

t2 - t - 2 = 0

t = { - (-1) ± SQRT[ ( - 1)2 - 4 (1) (-2)] }/2(1)

t = { 1 ± SQRT[ 1 + 8)] }/2(1)

t = { 1 ± SQRT[ 9] }/2

t = { 1 ± 3 }/2

t1 = - 2/2 = - 1 = - 1 s (extaneous; throw out)

t2 = + 4/2 = 2 = 2 s

t = 2 s

(b) What is its velocity just before impact with the ground!

v2 = vi2 + 2 a (s - si)

vy2 = vyi2 + 2 ay (y - yi)

vy2 = (5 m/s)2 + 2 ( - 10 m/s2) (0 - 10 m)

vy2 = 52 + 2 ( - 10) ( - 10 )

vy2 = 225

vy = 15 m/s

vy = 15 m/s

2. A ball is thrown from the floor of a canyon as shown in the sketch above. The ball is thrown with an initial speed of 30 m/s at an angle of 53° above the horizontal. It lands on the plateau above the canyon, 20 m above the canyon floor.

(a) How long is the ball in the air?

We need the y-component of the initial velocity,

vyi = vi sin = (30 m/s) sin 53o = (30 m/s) (0.8) = 24 m/s

For what time do we find y = 20 m?

y = yi + vyi t + (1/2) ay t2

y = 0 + (24 m/s) t + (0.5) ( - 10 m/s2) t2

Set y = 20 m and solve for the time t,

20 m = 0 + (24 m/s) t + (0.5) ( - 10 m/s2) t2

20 = 24 t - 5 t2

5 t2 - 24 t + 20 = 0

t = [ - ( - 24) ± SQRT { ( - 24 )^2 - 4 (5) (20) }] / [ 2 (5) ]

t = [ 24 ± SQRT { 576 - 400 } ] / 10

t = [ 24 ± SQRT { 176 } ] / 10

t = [ 24 ± 13.3 ] / 10

t1 = 37.3 / 10 = 3.73 sec

t2 = 10.7 / 10 = 1.07 sec

t = 1.07 s

The time t1 = 3.73 seconds is the time we want. At that time, the ball has gone up, reached its max height, and is on its way down as it strikes the plateau at y = 20 m. The time t2 = 1.07 seconds is the time -- on its way up -- when the ball passes through y = 20 m.

(b) Where does it strike the plateau, 20 m above the canyon floor? That is, how far has it traveled horizontally when it strikes the plateau?

Now we are ready to ask how far the ball travels horizontally during that time of t = 3.73 s. For that, of course, we need the horizontal velocity,

vx = vxi = vi cos = vi cos 53o = (30 m/s) (0.6) = 18 m/s

x = vx t = (18 m/s) (3.73 s)

x = 67.1 m

 3. Find the tension in the three ropes. Feel free to use the approximation of g = 10 m/s2.   Essentially by inspection, we can immediately set T3 = w = (10 kg) (10 m/s2) T3 = 100 N

Now we will concentrate on the "know" where the three ropes are joined. Since it is in equilibrium, the three forces must have a sum or resultant of zero.

Fnet = F = T1 + T2 + T3 = 0

Fnet,x = Fx = T1x + T2x + T3x = 0

T1x + T2x + T3x = - T1 cos 60o + T2+ 0 = 0

- T1 (0.500) + T2 = 0

0.500 T1 = T2

Fnet,y = Fy = T1y + T2y + T3y = 0

T1y + T2y + T3y = T1 sin 60o + 0 - 100 N = 0

T1 (0.866) = 100 N

T1 = (100 / 0.866) N

T1 = 115 N

T2 = 0.500 T1 = 0.500 (115 N)

T2 = 57.5 N

T3 = 100 N

4. Consider the "inclined Atwoods machine" sketched above.

The masses are m1 = 1 kg and m2 = 3 kg.

The plane is inclined at an angle of = 37o.

The coefficient of friction between mass m2 and the plane is 0.1

Find the acceleration of the masses.

 We expect the acceleration to be upward and have that drawn beside the free-body diagram. As always, we are now ready to apply F = m a to these forces acting on this object. Fnet = F = T - m1 g = m1 a T - m1 g = m1 a T - (1 kg) (10 m/s2) = (1 kg) a As we might expect by now, this one equation has two unknowns -- tension T and acceleration a -- so we must look elsewhere for additional information. Of course, the place to look is at the other mass.
Carefully construct a free-body diagram showing all the forces acting on mass m2.

There are four forces acting on this mass -- the string exerts a force T, the (frictionless) inclined plane exerts a "normal" force n, gravity pulls down with a force of w1 = m1 g, and there is the friction force Ff. To find the net force, we must resolve these vectors into their components. Since the acceleration will be along the direction of the plane, we have chosen that direction as the x-axis.

Notice that the angle in this diagram is measured from the y-axis. That means the weight has components of

wx = m2 g sin = (3 kg)(10 m/s2) sin 37o = (30 N) (0.6)

wx = 18 N

wy = - m2 g cos = (3 kg)(10 m/s2) cos 37o = (30 N) (0.8)

wx = 24 N

And we have

nx = 0

ny = n

That is, the normal force has only a component in the positive y-direction.

Ff,x = - Ff

Ff,y = 0

That is, the friction force has only a component in the negative x-direction.

and
Tx = - T

Ty = 0

That is, the rope exerts only a component in the negative x-direction.

Now we can apply Newton's Second Law to this mass:

F = m a

F = Fnet = T + n + w + Ff = m a

Fx = Fnet,x = Fx = m2 ax

Fx = Tx + n x + w x + Ff,x = m2 ax

- T + 0 + 24 N - Ff = m2 ax = (3 kg) a

where we have used

ax = a

since the acceleration is only in the positive x-direction.

- T + 24 N - Ff = (3 kg) a

But we do not yet know the friction force. We know only

Ff = n

Ff = 0.1 n

So we must find n, the normal force. We can do this by looking at the sum of the forces in the y-direction,

Fy = Ty + n y + w y + Ff,y = m2 ay = 0

Ty + n y + w y + Ff,y = 0

0 + ny - 18 N + 0 = 0

ny = 18 N

Therefore

Ff = (0.1) (18 N)

Ff = 1.8 N

Now,

- T + 24 N - Ff = (3 kg) a

becomes

- T + 24 N - 1.8 N = (3 kg) a

- T + 22.2 N = (3 kg) a

T = 22.2 N - (3 kg) a

Along with the equation from the hanging mass m1, this provides all the information we need to solve for T and a. As before, we can solve one of these equations for T and substitute that into the other equation, from the hanging mass m1, and solve for a.

T = 22.2 N - (3 kg) a

T - (1 kg) (10 m/s2) = (1 kg) a

[ 22.2 N - (3 kg) a ] - (1 kg) (10 m/s2) = (1 kg) a

22.2 N - (3 kg) a - 10 N = (1 kg) a

12.2 N = (4 kg) a

a = (12.2/4) (N/kg)

a = 3.05 m/s2