Prof ML Lekala Department of Physics University University of South Afric a 2011
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PHY1501/may-june/memo/2011
SECTION B (WRITTEN SOLUTIONS)
QUESTION 1 (Definitions)
[10 marks]
a) Conservative force: A force is conservative when the work it does o n a moving object is independent of the path between the object’s initial and final positions. OR. A force is conservative when it does no net work on an object moving around a closed path, starting and finishing at the same point. b) Stress: Stress is the magnitude of the force per unit area applied to an object and causes strain. (stress = ) c) Torque: Torque is the product of the magnitude of the applied force and the lever arm ( ). d) Center of gravity: The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated. e) Center of mass: The center of mass is a point that represents the average location for the total mass of a system.
QUESTION 2 (Principles)
[8 marks]
a) Archimedes’ principle: Any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals to the weight of the fluid that the object displaces, i.e. b) Bernoulli’s principle: In the steady flow of a nonviscous, incompressible fluid of density pressure fluid speed and the elevation at any two point (1 and 2) are related by . c) Principle of conservation of angular momentum: The total angular momentum of a system remains constant if the net average external torque acting on the system is zero. d) Principle of conservation of total mechanical energy : The total mechanical energy (E = KE+PE) of an object remains constant as the object moves, provided that the net work done by external nonconservative forces is zero.
QUESTION 3
[10 marks]
We first decompose the three coplanar forces into their rectangular components, as shown in the table below. F1
F2
x-component
-400 N
y-component
N
F3 (250 N)sin 45° = 176.8 N (250 N)cos 45° = 176.8 N
-(200 N)(4/5) = -160 N (200 N)(3/5) = 120 N
In order to determine the magnitude of the resultant force we first sum the forces in the x- and ydirection, and then use the Pythagoras theorem to determine the resultant. Therefore
∑ F x = −400N + 176.8N −160N = − 383.2 N ∑ F y = 0N + 176.8N + 120N=296.8 N The negative sign indicates
F R = (
acts to the left. The magnitude of the resultant is
∑F ) x
2
+ (∑ F y ) 2 = ( −383.2N) 2 + (296.8N) 2 = 485 N
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PHY1501/may-june/memo/2011
The direction of
is given by θ =
∑ F y
tan −1
= tan −1 296.8N −383.2N = −37.8° ∑ F x
(
)
i.e. 34.9° below the x-axis. Pictorially
y
θ
x
QUESTION 4
[8 marks]
Suppose the board is kept from rotating and rests on the support as shown in the picture. The net torque due to the weights of the two blocks is τ
= W12 kg 12 kg − W 4 kg 4kg = (12kg) g (0.60m) − (4.0kg) g (1.4m) = [(12kg)(0.60m) − (4.0kg)(1.4m)](9.80m/s2 ) = 15.68 N m
When the system is allowed to rotate by Newton’s second law of motion for rotational motion, , where is the moment of inertia and is the magnitude of the angular acceleration. Therefore we have α =
τ
I
=
15.68 N 12 kg m
= 1.3 rad/s
2
2
QUESTION 5
[10 marks]
The conservation of energy applied between point A and the top of the trajectory gives KE A + mgh A = mgh where h = 4.00 m. Rearranging, we find KE A = mg (h – h A ) or
vA =
(
2 g ( h − hA ) = 2 9.80 m/s
2
) ( 4.00 m − 3.00 m ) =
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4.43 m/s
PHY1501/may-june/memo/2011
QUESTION 6
[5 marks]
Let the constant angular acceleration of the object be α, and the radial distance of a point on the object. Since the object starts from rest the initial angular velocity is zero. Hence from we have . But Thus and are constants. Therefore for different times we have
ac (3s) = (rα 2 )(3s) 2 and ac (6s) = ( rα 2 )(6s) 2 Dividing the two equations by each other we obtai n a
(6s)
a
(3s)
c
c
=
( r α 2 )(6s)2 (r α 2 )(3s)2
=
4
Hence magnitude of the centripetal acceleration of the point six seconds after the motion had begun is
ac (6s) = 4 × ac (3s)= 4(2.0m/s 2 ) = 8 m/s 2 QUESTION 7
[4 marks]
Method 1: The work done by a constant force is given by where is the magnitude of the average force, is the magnitude of the displacement ( ), and is the angle between the force and the displacement. The force acting on the spring is given by and as spring contracts the magnitude of this force changes from to Therefore the magnitude of the average force is . Hence the work done by the average spring force i s
Welastic = ( Fav cos θ ) s = = Method 2: The work done by a force
1 2
kx02 −
1 2
1 2
( kx0 + kx f )(cos 0°)( x0 − x f )
kx f 2
in compressing the spring from x f
Welastic =
x f
1
∫ Fdx = −k ∫ xd x = −k 2 x
x0
x0
1
1
2
2
= kx02 − kx f 2
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to
is given by x f
2
x
0
PHY1501/may-june/memo/2011 QUESTION 8 Consider the diagram in the question. Since the structure is in equilibrium the net force (in component form) is zero; and • • the net torque is zero Axes: Choose down and left as (-ve) and up and right as (+ve): Table of components of the forces component 400 N
xy-
400 N 0
200 N (200 N)cos -
800 N 0 -800 N
1000 N 0 -1000 N
500 N -(500 N)sin
600 N -
F 0 F
Hence a) Taking moments about point A to obtain magnitude of force F, we obtain
F = .... = 926, 8N b) Summing components of the forces in the x- and y-direction to zero, we obtain reaction force at A (R A) c) R A
