Question 1

A point  A  on a sphere of weight  W  rests in contact with a smooth vertical wall and is supported by a string joining a point  B  on the sphere to a point  C  on the wall. Draw a free body diagram of the sphere.

Accepted Answer

Weight  (W)  of the sphere acts in the downward direction, rope exerts a tension force  (T)  along its length and the wall exerts a normal force  (N)  on the sphere as shown in the figure.

Question 2

A rod OA is suspended with help of a massless string AB as shown in figure. Rod is hinged at point O. Draw free body diagram of the rod.

Accepted Answer

\displaystyle N_{1} =normal reaction between sphere and wall, \displaystyle N_{2} =normal reaction between sphere and ground \displaystyle N_{3} =normal reaction between sphere and road and \displaystyle N_{4} =normal reaction beween rod and ground f= force of friction between rod and ground

Question 3

A helicopter is moving to the right at a constant horizontal velocity. It experience three forces  \ { \overrightarrow { F }  }_{ gravitational } ,  \ { \overrightarrow { F }  }_{ drag }  and force on it caused by rotor  \ { \overrightarrow { F }  }_{ rotor } . Which of the following diagrams can be a correct free-body diagram representing force on the helicopter?

Accepted Answer

It is given that the helicopter is moving with constant horizontal velocity  (a=0)  1. We know that drag force is always opposite to the motion.2. Rotor Force is perpendicular to the plane of rotation of rotors.3. And the gravity acts downwards.Hence Option C is  correct

Question 4

A  5.00kg  block is placed on top of a  10.0kg  block (Fig. above). A horizontal force of  45.0 N  is applied to the  10kg  block, and the  5.00kg  block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is  0.200 . (a) Draw a free-body diagram for each block and identify the actionreaction forces between the blocks. (b) Determine the tension in the string and the magnitude of the acceleration of the  10.0kg  block.

Accepted Answer

(a) The free-body diagrams are shown in the figure below.     f_{1}  and  n_{1}  appear in both diagrams as action-reaction pairs.(b) For the  5.00kg  mass, Newton&#8217;s second law in the  y  direction gives:     n_{1} = m_{1}g = (5.00 kg)(9.80 m/s^{2} ) = 49.0 N In the  x  direction,     f_{1} &#8722; T = 0      T = f_{1} = \mu mg = 0.200(5.00 kg)(9.80 m/s^{2} ) = 9.80 N For the  10.0kg  mass, Newton&#8217;s second law in the  x  direction gives:     45.0 N &#8722; f_{1} &#8722; f_{2} = (10.0 kg)a In the  y  direction,     n_{2} &#8722; n_{1} &#8722; 98.0 N = 0      f_{2} = \mu n_{2} = \mu (n_{1} + 98.0 N) = 0.20(49.0 N + 98.0 N) = 29.4 N      45.0 N &#8722; 9.80 N &#8722; 29.4 N = (10.0 kg)a      a = 0.580 m/s^{2}

Question 5

Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (Fig. above). Suppose

F = 68.0 N, m_{1} = 12.0 k g, m_{2} = 18.0 k g

, and the coefficient of kinetic friction between each block and the surface is

0.100

. (a) Draw a free-body diagram for each block. Determine (b) the acceleration of the system and (c) the tension

T

in the rope.

Accepted Answer

Because the cord has constant length, both blocks move the same number of centimeters in each second and so move with the same acceleration. To find just this acceleration, we could model the  30kg  system as a particle under a net force. That method would not help to finding the tension,so we treat the two blocks as separate accelerating particles.(a) Below figure shows the free-body diagrams for the two blocks.The tension force exerted by block  1  on block  2  is the same size as the tension force exerted by object  2  on object  1 . The tension in a light string is a constant along its length, and tells how strongly the string pulls on objects at both ends.(b) We use the free-body diagrams to apply Newton&#8217;s second law.For  m_{1}: \sum F_{x} = T &#8722; f_{1} = m_{1}a  or  T = m_{1}a + f_{1} ........... [1]And also     \sum F_{y} = n_{1} &#8722; m_{1} g = 0  or  n_{1} = m_{1}g Also, the definition of the coefficient of friction gives     f_{1} = \mu n_{1} = (0.100)(12.0 kg)(9.80 m/s^{2}) = 11.8 N For  m_{2}: \sum F_{x} = F &#8211; T &#8211; f_{2} = ma ...................................... [2]Also from the  y  component,  n_{2} &#8211; m_{2}g = 0  or   n_{2} = m_{2}g And again   f_{2} = \mu n_{2} = (0.100)(18.0 kg)(9.80 m/s^{2}) = 17.6 N Substituting  T  from equation [1] into [2], we get     F &#8722; m_{1}a &#8722; f_{1} &#8722; f_{2} = m_{2}a  or  F &#8722; f_{1} &#8722; f_{2} = m_{2}a + m_{1}a Solvind for  a      a=\frac{F-f_{1}-f_{2}}{m_{1}+m_{2}}=\frac{(68.0N-11.8N-17.6N)}{(12.0kg+18.0kg)}=1.29m/s^{2} (c) From equation [1]     T=m_{1}a+f_{1}=(12.0kg)(1.29m/s^{2})+11.8N=27.2N

Question 6

Three objects are connected on a table as shown in above figure. The coefficient of kinetic friction between the block of mass

m_{2}

and the table is

0.350

. The objects have masses of

m_{1} = 4.00 k g, m_{2} = 1.00 k g

, and

m_{3} = 2.00 k g

and the pulleys are frictionless. (a) Draw a free body diagram of each object. (b) Determine the acceleration of each object, including its direction. (c) Determine the tensions in the two cords. What If? (d) If the tabletop were smooth, would the tensions increase, decrease, or remain the same? Explain.

Accepted Answer

(a) The free-body diagrams for each object are given below.(b) Let  a  represent the positive magnitude of the acceleration  &#8722;a\hat{j}  of  m_{1} , of the acceleration  &#8722;a\hat{i}  of  m_{2} , and of the acceleration  +a\hat{j}  of  m_{3} . Call  T_{12}  the tension in the left cord and  T_{23}  the tension in the cord on the right.For  m_{1}, \sum F_{y} = ma_{y} : +T_{12} &#8722; m_{1}g = &#8722;m_{1}a For  m_{2}, \sum F_{x} = ma_{x} : &#8722;T_{12} + \mu_{k}n + T_{23} = &#8722;m_{2}a and   \sum F_{y} = ma_{y} , giving   n &#8722;m_{2}g = 0 .For   m_{3}, \sum F_{y} = ma_{y} , giving   T_{23} &#8722;m_{3}g = +m_{3}a .We have three simultaneous equations:                 &#8722;T_{12} + 39.2 N = (4.00 kg)a +T_{12} &#8722; 0.350(9.80 N) &#8722; T_{23} = (1.00 kg)a                  +T_{23} &#8722; 19.6 N = (2.00 kg)a Add them up (this cancels out the tensions):             +39.2 N &#8722; 3.43 N &#8722; 19.6 N = (7.00 kg)a              a = 2.31 m/s^{2} , down for  m_{1} , left for  m_{2} , and up for  m_{3} (c) Now    &#8722;T_{12} + 39.2 N = (4.00 kg)(2.31 m/s^{2} )            T_{12} = 30.0 N and          T_{23} &#8722; 19.6 N = (2.00 kg)(2.31 m/s^{2} )           T_{23} = 24.2 N (d) If the tabletop were smooth, friction disappears  (\mu_{k} = 0) , and so the acceleration would become larger. For a larger acceleration, according to the equations above, the tensions change:          T_{12} = m_{1}g &#8722; m_{1}a\rightarrow  T_{12}  decreases.          T_{23} = m_{3}g + m_{3}a\rightarrow  T_{23}  increases.

Question 7

A woman at an airport is towing her  20.0kg  suitcase at constant speed by pulling on a strap at an angle  	heta  above the horizontal (Fig. above). She pulls on the strap with a  35.0N  force, and the friction force on the suitcase is  20.0 N . (a) Draw a free body diagram of the suitcase. (b) What angle does the strap make with the horizontal? (c) What is the magnitude of the normal force that the ground exerts on the suitcase?

Accepted Answer

(a) See the free-body diagram of the suitcase in figure below.(b)  m_{suitcase} = 20.0 kg, F = 35.0 N      \sum F_{x}  = ma_{x} : &#8722;20.0 N + F \cos&#952; = 0      \sum F_{y}  = ma_{y} : +n + F \sin&#952; &#8722; F_{g} = 0      F \cos&#952; = 20.0 N          \cos	heta=\frac{20.0N}{35.0N}=0.571          	heta=55.2^{0} (c) With  F_{g} = (20.0 kg)(9.80 m/s^{2}) ,          n = F_{g} &#8722; F \sin&#952; = [196 N&#8722;(35.0 N)(0.821)]           n = 167 N

Question 8

Mandy stands on a weighing scale inside a lift (elevator) that accelerates vertically upwards as shown in the diagram below. The forces on Mandy are her weight W and the reaction force from the scale R. The reading of scale is:

Accepted Answer

The reading of the scale will show reaction force  R . But the normal force will be more than the weight of the body as it is accelerating upward.

Question 9

The apparent weight of a person in a lift moving downwards is half his apparent weight in the same lift moving upwards with the same magnitude of acceleration. Acceleration of the lift is:

Accepted Answer

Apparent weight of a person in lift will be  m(g+a)  while moving upwards whereas it will be  m(g-a)  while moving downwards.Hence, as per the given information, m(g+a)=2m(g-a) g+a=2g-2a Hence,  a=\dfrac { g }{ 3 }

Laws Of Motion