The mass of the block is 1.8 107 kg, the dip angle è of the bedding plane is 26°, and the coeffient of static friction between block and plane is 0.63.
(a) Show that the block will not slide.
(b) Water seeps into the joint and expands upon freezing, exerting on the block a force F parallel to AA'. What minimum value of F will trigger a slide?
To determine whether the block will slide or not, we need to compare the force of gravity acting on the block and the maximum static friction force that can be exerted between the block and the plane. If the force of gravity is greater than the maximum static friction force, the block will slide.
(a) Let's calculate the maximum static friction force first. The formula for the maximum static friction force is given by:
f_max = µ × N,
where f_max is the maximum static friction force, µ is the coefficient of static friction, and N is the normal force exerted on the block by the plane.
The normal force can be determined by decomposing the weight of the block into its components. The weight of the block can be calculated using the formula:
W = m × g,
where W is the weight, m is the mass of the block, and g is the acceleration due to gravity.
In this case, the mass of the block is given as 1.8 × 10^7 kg.
W = (1.8 × 10^7 kg) × (9.8 m/s^2) = 1.764 × 10^8 N.
Next, we need to calculate the normal force. The normal force is equal to the component of the weight perpendicular to the plane, which can be calculated as:
N = W × cos(è),
where è is the dip angle of the bedding plane.
Substituting the values, we have:
N = (1.764 × 10^8 N) × cos(26°) = 1.586 × 10^8 N.
Finally, we can calculate the maximum static friction force:
f_max = µ × N = 0.63 × (1.586 × 10^8 N) = 9.986 × 10^7 N.
Now, we compare f_max with the force of gravity acting on the block (W). We have f_max = 9.986 × 10^7 N and W = 1.764 × 10^8 N. Since f_max is less than W, it means that the block will not slide. Hence, we can conclude that the block will remain at rest.
(b) To determine the minimum value of the force F that will trigger a slide, we need to compare it with the maximum static friction force (f_max) we calculated above.
The force F required to trigger the slide should be equal to or greater than f_max. Therefore, the minimum value of F can be evaluated as:
F_min = f_max = 9.986 × 10^7 N.
So, a force of at least 9.986 × 10^7 N exerted parallel to AA' will be required to trigger a slide.