A block P of mass 2.75 kg is lying on a rough inclined plane of angle è=0.87 radians. Block P is attached via a model string that passes over a model pulley to a second block Q of unknown mass M kg hanging vertically. The coefficient of static friction between the block P and the inclined plane is ì=0.39.
Calculate the value of the mass of block Q, in kg, that is just sufficient to initiate the sliding motion of block P up the inclined plane.
Thanks!
(360Deg/6.28rad) * 0.87rad = 49.9 Deg.
Wb = mg = 2.75kg * 9.8N/kg = 26.95 N. =
Wt. of block P.
Fb = 26.95N @ 49.9 Deg.
Fp = 26.95*sin49.9 = 20.61 N. = Force
parallel to incline.
Fv = 26.95*cos49.9 = 17.36 N. = Force
perpendicular to incline.
Fn = Fb2 - Fp - Fs = 0.
Fb2 - 20.61 - 0.39*17.36 = 0.
Fb2 - 27.38 = 0.
Fb2 = 27.38 N. = Force of block Q.
mg = Fb2 = 27.38.
m = 27.38/g = 27.38/9.8 = 2.79 kg = Mass 0f block Q.
To determine the value of the mass of block Q that is just sufficient to initiate the sliding motion of block P up the inclined plane, we need to analyze the forces acting on block P.
1. Determine the normal force (N) acting on block P:
The normal force is equal to the perpendicular component of the weight of block P, which is given by N = mg * cos(θ), where m is the mass of P and θ is the angle of the inclined plane.
Substituting the given values, we have N = (2.75 kg) * 9.8 m/s^2 * cos(0.87 radians).
2. Determine the maximum force of static friction (F_max) between block P and the inclined plane:
The maximum force of static friction can be calculated using the equation F_max = μ * N, where μ is the coefficient of static friction.
Substituting the given value, we have F_max = 0.39 * N.
3. Determine the force due to the weight of block P (W) acting along the inclined plane:
The weight is given by W = mg, where m is the mass of P.
Substituting the given value, we have W = (2.75 kg) * 9.8 m/s^2.
4. Set up an equilibrium equation for the forces along the inclined plane:
Taking the x-direction as parallel to the inclined plane, we have:
F_max = W * sin(θ).
5. Substitute the previously derived expressions for F_max and W:
0.39 * N = (2.75 kg) * 9.8 m/s^2 * sin(0.87 radians).
6. Solve for N using the expression from step 1:
N = (2.75 kg) * 9.8 m/s^2 * cos(0.87 radians).
7. Substitute the expression for N in step 6 into the equation in step 5:
0.39 * (2.75 kg) * 9.8 m/s^2 * cos(0.87 radians) = (2.75 kg) * 9.8 m/s^2 * sin(0.87 radians).
8. Simplify and solve for M, the mass of block Q:
M = (0.39 * cos(0.87 radians)) / sin(0.87 radians).
Using a calculator, compute the value of M to find the mass of block Q that is just sufficient to initiate the sliding motion of block P up the inclined plane.