21.
Like #20, this problem can most easily be solved using separate free-body analyses on the individual masses. Free-body 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 y-axis.
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 x-axis.)
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 free-body 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 re-written 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 re-inserted 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