[ Questions #1#12  Questions #13#21  Questions #22#30 ]

Use the approximation that g= ~10 m/s^{2} to fill in the blanks in the following diagrams.
13. F_{grav} = m•g = ~800 N ∑F_{y} = ma_{y} = (80 kg)•(2.0 m/s/s) ∑F_{y} = 160 N, down The F_{grav} (down) and the F_{air} (up) must add up to 160 N, down. Thus, F_{air} must be smaller than F_{grav} by 160 N. F_{air} = 640 N 
14. Since F_{grav} = m•g, m can be calculated to be ~70 kg (m=F_{grav}/g). Since a_{y} = 0 m/s/s, F_{norm} must equal F_{grav}; so F_{norm} = 700 N. ∑F_{x} = m•a_{x} = (70 kg)•(5.0 m/s/s) ∑F_{x} = 350 N, left (Note that the ·F_{x} direction is always the same as the a_{x} direction.) With F_{app} being the only horizontal force, its value must be 350 N  the same as ·F_{x}. 
15. F_{grav} = m•g = ~800 N Since there are two forces pulling upwards and since the sign is hanging symmetrically, each force must supply an upwards pull equal to onehalf the object's weight. So the vert pull (F_{y}) in each force is 400 N. The following triangle can be set up: Using trig, we can write: sin(30 deg.)=(400 N)/F_{tens} Solving for F_{tens} yields 800 N. 
16. A quick blank is F_{grav}: F_{grav} = m•g = ~800 N Now resolve the 60N force into components using trigonometry and the given angle measure: F_{x} = 60 N•cos(30 deg) = 52 N F_{y} = 60 N•sin(30 deg) = 25 N Since the acceleration is horizontal, the sum of the vertical forces must equal 0 N. So F_{grav} = F_{y} + F_{norm}. Therefore F_{norm} = F_{grav}  F_{y} = 50 N. Knowing F_{norm} and mu, the F_{frict} can be determined: F_{frict}_{ }= mu•F_{norm} = 0.5*(50 N) = 25 N Now the horizontal forces can be summed: ∑F_{x} = F_{x} + F_{frict} = 52 N, right + 25 N, left ∑F_{x }= 27 N, right Using Newton's second law, ∑F_{x} = m•a_{x} So a_{x }= (27 N)/(8 kg) = 3.8 m/s/s, right 
17. The first step in an inclined plane problem is to resolve the weight vector into parallel and perpendicular components: F_{par} = m•g•sin(angle) = (420 N)•sin(30 deg) = 210 N F_{perp} = m•g•cos(angle) = (420 N)•cos(30 deg) = 364 N The mass can be found as m = F_{grav}/g m = F_{grav}/g = (420 N)/(10 m/s/s) = ~42 kg The F_{norm} acts opposite of and balances the F_{perp}. So F_{norm} = F_{perp} = 364 N Knowing F_{norm} and mu, the F_{frict} can be determined: F_{frict}_{ }= mu•F_{norm} = 0.2*(364 N) = 73 N Now the forces parallel to the incline can be summed: ∑F_{} = F_{} + F_{frict} = 210, down to left + 73 N, up to right ∑F_{ }= 137 N, down to left Using Newton's second law, ∑F_{} = m•a_{} So a_{}_{ }= (137 N)/(42 kg) = 3.3 m/s/s 
Treating the two masses as a single system, it can be concluded that the net force on the 9kg system is: ∑F_{system} = m•a_{system} = (9 kg)•(2.5 m/s/s) = 22.5 N, right The freebody diagram for the system is: The F_{norm} supporting the 9kg system is ~90 N. So the F_{frict} acting upon the system is: F_{frict} = mu• F_{norm} = 0.20*(90 N) = 18 N, left So if ∑F_{system }= 22.5 N, right and F_{frict} = 18 N, left, the rightward F_{tens1} must equal 40.5 N. The F_{tens2} force is found inside the system; as such it can not be determined through a system analysis. To determine the F_{tens2}, one of the individual masses must be isolated and a freebody analysis must be conducted for it. The 3kg mass is selected and analyzed: The F_{norm} and F_{grav} balance each other; their value is ~30 N. The F_{frict} on the 3kg mass is: F_{frict} = mu•F_{norm} = 0.20*(30 N) = 6 N, left The net force on the 3kg object is: ∑F_{x} = m•a_{x} = (3 kg)•(2.5 m/s/s) = 7.5 N, right The horizontal forces must sum up to the net force on the 3kg object; So ∑F_{x} = m•a_{x} = F_{tens2} (right) + F_{frict} (left) 7.5 N, right = F_{tens2} + 6 N, left The F_{tens2} must be 13.5 N. 
19. Like most twobody problems involving pulleys, it is usually easiest to forgo the system analysis and conduct separate freebody analyses on the individual masses. Freebody diagrams, the chosen axes systems, and associated information is shown below. Analyzing the F_{x} forces on the 250g mass yields: _{}ma_{x} = F_{tens}  F_{frict} Since F_{frict} = mu•F_{norm} and F_{norm} = 2.5 N The F_{frict} is (0.1)*(2.5 N) = 0.25 N. Substituting into equation 1 yields (0.250 kg)•a_{x} = F_{tens}  0.25 N
Analyzing the F_{y} forces on the 50g mass yields: _{}ma_{y} = F_{grav}  F_{tens} Substituting m and F_{grav} values into equation 3 yeilds: (0.050 kg)*a_{y} = (0.500 N)  F_{tens} The above equation can be rearranged to: F_{tens} = (0.500 N)  (0.050 kg)*a_{y} Equation 4 provides an expression for Ftens; this can be substituted into equation 2: (0.250 kg)•a_{x} = (0.500 N)  (0.050 kg)*a_{y}  0.25 N Now since both masses accelerate at the same rate, a_{x }=a_{y} and the above equation can be simplified into an equation with 1 unknown  the acceleration (a): (0.250 kg)•a = (0.500 N)  (0.050 kg)*a  0.25 N After a few algebra steps, the acceleration can be found: (0.0300 kg)•a = 0.25 N a = 0.833 m/s/s Now that a has been found, its value can be substituted back into equation 4 in order to solve for F_{tens}: F_{tens} = (0.500 N)  (0.050 kg)*(0.833 m/s/s) F_{tens} = 0.458 N 
This problem can most easily be solved using separate freebody analyses on the individual masses. Freebody diagrams, the chosen axes systems, and associated information is shown below. Note that the positive yaxis is chosen as being downards on the 200g mass since that is the direction of its acceleration. Similarly, it chosen as upwards on the 100g mass since that is the direction of its acceleration. For the 200gram mass, the sum of the vertical forces equals the mass times the acceleration: F_{grav}  F_{tens} = m•a_{y} 2.00 N  F_{tens} = (0.200 kg)•a_{y} The same type of analysis can be conducted for the 100gram mass: F_{tens}  F_{grav} = m•a_{y} F_{tens}  1.00 N = (0.100 kg)•a_{y} Equation 2 can be rearranged to obtain an expression for the tension force: F_{tens} = (0.100 kg)•a_{y} + 1.00 N This expression for F_{tens} can be substituted into equation 1 in order to obtain a single equation with acceleration (a_{y}) as the unknown. The a_{y} value can be solved for. 2.00 N [(0.100 kg)•a_{y} + 1.00 N] = (0.200 kg)•a_{y} 2.00 N  1.00 N = (0.200 kg)•a_{y} + (0.100 kg)•a_{y} 1.00 N = (0.300 kg)•a_{y} a_{y} = (1.00 N)/(0.300 kg) = 3.33 m/s/s Now with a_{y} known, its value can be substituted into equation 3 in order to determine the tension force: F_{tens} = (0.100 kg)•a_{y} + 1.00 N F_{tens} = (0.100 kg)•(3.33 m/s/s) + 1.00 N F_{tens} = 0.333 N + 1.00 N = 1.33 N 
21. Like #20, this problem can most easily be solved using separate freebody analyses on the individual masses. Freebody diagrams, the chosen axes systems, and associated information is shown below. Note that in chosing the axis system, it has been assumed that object 1 will accelerate up the hill and object 2 will accelerate downwards. If this ends up to be false, then the acceleration values will turn out to be negative values. Object 1 is on an inclined plane. The usual circumstances apply; their is no acceleration along what has been designated as the yaxis. F_{norm} = F_{perp} = m•g•cos(theta) = 888.2 N The parallel component of F_{grav} is F_{} = m•g•sin(theta) = (100 kg)•(9.8 m/s^{2})•sin(25) F_{} = 414.2 N The F_{frict} value can be found from the F_{norm} value: F_{frict} = mu•F_{norm} = (0.35)•(888.2 N) = 310.9 N The ∑F_{x} = m•a_{x} equation can now be written: ∑F_{x} = m•a_{x} F_{tens}  F_{frict } F_{} = m•a_{x} F_{tens}  310.9 N_{ } 414.2 N = m•a_{x} (Note that the F_{frict} and F_{} forces are subtracted from F_{tens} since they are heading in the direction of the negative xaxis.) The above process can be repeated for object 2. The ∑F_{y} = m•a_{y} equation can now be written: ∑F_{y} = m•a_{y} F_{grav}  F_{tens } = m•a_{y} (980 N)  F_{tens } = m•a_{y} The separate freebody analyses have provided two equations with two unknowns; the task at hand is to use these two equations to solve for F_{tens} and a. Equation 2 can be rewritten as (980 N)  m•a_{y} = F_{tens} Since both objects accelerate together at the same rate, the ax for object 1 is equal to the ay value for object 2. The subscripts x and y can be dropped and a can be inserted into each equation. (980 N)  m•a = F_{tens} Equation 3 provides an expression for F_{tens} in terms of a. This expression is inserted into equation 1 in order to solve for acceleration. The steps are shown below. (980 N)  m•a  310.9 N_{ } 414.2 N = m•a = 2•m•a 254.9 N = 2•(100 kg)•a 1.27 m/s^{2} = a The value of a can be reinserted into equation 3 in order to solve for F_{tens}: F_{tens} = (980 N)  m•a =(980 N)  (100 kg)•(1.27 m/s^{2}) F_{tens} = 853 N 

Finding Acceleration  Net Force Problems Revisited  Inclined Planes 
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