Problem #1
Find the tension in each cord for the systems shown in Figure. (Neglect the mass of the cords.)
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| Fig.1 |
Answer:
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| Fig.2 |
(a) In this part, we solve the system shown in Figure 1(a). Think of the forces acting on the 5.0 kg mass (which we’ll call m1). Gravity pulls downward with a force of magnitude mg. The vertical string pulls upward with a force of magnitude T3. (These forces are shown in Fig. 1(a).) Since the hanging mass has no acceleration, it must be true that T3 = m1g. This gives us the value of T3:
T3 = m1g = (5.0kg)(9.80 m/s2) = 49N
Next we look at the forces which act at the point where all three strings join; these are shown in Fig.2(b). The force which the strings exert all point outward from the joining point and from simple geometry they have the directions shown Now this point is not accelerating either, so the forces on it must all sum to zero. The horizontal components and the vertical components of these forces separately sum to zero.
The horizontal components give:
−T1 cos 400 + T2 cos 500 = 0
This equation by itself does not let us solve for the tensions, but it does give us:
T2 cos 500 = T1 cos 400 ⇒ T2 = [cos 400/cos500]/T2 = 1.19T1
The vertical forces sum to zero, and this gives us:
T1 sin 400 + T2 sin 500 −T3 = 0
We already know the value of T3. Substituting this and also the expression for T2 which we just found, we get:
T1 sin 400 + (1.19T1) sin 500 – 49 N = 0
and now we can solve for T1. A little rearranging gives:
(1.56)T1 = 49 N
which gives
T1 = 49 N/(1.56) = 31.5 N
Now with T1 in hand we get T2:
T2 = (1.19)T1 = (1.19)(31.5 N) = 37.5 N
Summarizing, the tensions in the three strings for this part of the problem are
T1 = 31.5 N
T2 = 37.5 N
T3 = 49 N
(b) Now we study the system shown in Fig. 1(b). Once again, the net force on the hanging mass (which we call m2) must be zero. Since gravity pulls down with a a force m2g and the vertical string pulls upward with a force T3, we know that we just have T3 = m2g, so:
T3 = m2g = (10kg)(9.80 m/s2) = 98 N
Now consider the forces which act at the place where all the strings meet. We do as in part (a); the horizontal forces sum to zero, and this gives:
−T1 cos 600 + T2 = 0 ⇒ T2 = T1 cos 600
The vertical forces sum to zero, giving us:
T1 sin 600 − T3 = 0
But notice that since we know T3, this equation has only one unknown. We find:
T1 = T3 sin 600 = 98 N sin 600 = 113 N
Using this is our expression for T2 gives:
T2 = T1 cos 600 = (113 N) cos 600 = 56.6 N
Summarizing, the tensions in the three strings for this part of the problem are
T1 = 113 N
T2 = 56.6 N
T3 = 98 N
Problem #2
A 210kg motorcycle accelerates from 0 to 55 mi hr in 6.0s. (a) What is the magnitude of the motorcycle’s constant acceleration? (b) What is the magnitude of the net force causing the acceleration?
Answer:
(a) First, let’s convert some units:
55 mi/hr = (55 mi/hr)(1609 m/1 mi)(1 hr/3600s)= 24.6 m/s
so that the acceleration of the motorcycle is
a = (vx − vx0)/t = (24.6 m/s – 0)/6.0 s = 4.1 m/s2
(b) Now that we know the acceleration of the motorcycle (and its mass) we know the net horizontal force, because Newton’s Law tells us:
∑Fx = max = (210 kg)(4.1 m/s2) = 8.6 × 102 N
The magnitude of the net force on the motorcycle is 8.6 × 102 N.
Problem #3
A rocket and its payload have a total mass of 5.0 × 104 kg. How large is the force produced by the engine (the thrust) when
(a) the rocket is “hovering” over the launchpad just after ignition, and
(b) when the rocket is accelerating upward at 20 m/s2?
Answer:
(a) First thing: draw a diagram of the forces acting on the rocket! This is done in Figure. If the mass of the rocket is M then we know that gravity will be exerting a force Mg downward. The engines (actually, the gas rushing out of the rocket) exerts a force of magnitude Fthrust upward on the rocket.
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| Fig 3: (Forces acting on the rocket in Problem #3) |
If the rocket is hovering, i.e. it is motionless but off the ground then it has no acceleration; so, here, ay = 0. Newton’s Second Law then says:
∑Fy = Fthrust − Mg = May = 0
which gives
Fthrust = Mg = (5.0 × 104 kg)(9.80 m/s2) = 4.9 × 105 N
The engines exert an upward force of 4.9 × 105 N on the rocket.
(b) As in part (a), gravity and thrust are the only forces acting on the rocket, but now it has an acceleration of ay = 20 m/s2. So Newton’s Second Law now gives
∑Fy = Fthrust − Mg = May
so that the force of the engines is
Fthrust = Mg + May
= M(g + ay)
Fthrust = (5.0 × 104 kg)(9.80 m/s2 + 20 m/s2) = 1.5 × 106 N
Problem #4
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| Fig.4: (a) Block held in place on a smooth ramp by a horizontal force. (b) Forces acting on the block. |
A block of mass m = 2.0 kg is held in equilibrium on an incline of angle θ = 600 by the horizontal force F, as shown in Fig. (a).
(a) Determine the value of F, the magnitude of F.
(b) Determine the normal force exerted by the incline on the block (ignore friction).
Answer:
(a) The first thing to do is to draw a diagram of the forces acting on the block, which we do in Fig. 4(b). Gravity pulls downward with a force mg. The applied force, of magnitude F, is horizontal. The surface exerts a normal force N on the block; using a little geometry, we see that if the angle of the incline is 60◦, then the normal force is directed at 300 above the horizontal, as shown in Fig.4(b). There is no friction force from the surface, so we have shown all the forces acting on the block. Oftentimes for problems involving a block on a slope it is easiest to use the components of the gravity force along the slope and perpendicular to it. For this problem, this would not make things any easier since there is no motion along the slope. Now, the block is in equilibrium, meaning that it has no acceleration and the forces sum to zero. The fact that the vertical components of the forces sum to zero gives us:
N sin300 − mg = 0 ⇒ N = mg sin 300
Substitute and get:
N = (2.0 kg)(9.80 m/s2)/sin 300 = 39.2 N
The horizontal forces also sum to zero, giving:
F − N cos 300 = 0 ⇒ F = N cos 300 = (39.2 N) cos 300 = 33.9 N.
The applied force F is 33.9 N. (b) The magnitude of the normal force was found in part (a); there we found:
N = 39.2 N
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