Why do we use days and hours instead of hours?
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This question is incomplete, the complete question;
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.
Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb
Answer:
- the magnitude of force at point A is 79.1033 lb
- since FA < FA_max; Ladder WILL NOT slip
Explanation:
Given that;
∑'MA = 0
⇒ NB [Lsin∅] - W[L/2.cos∅] = 0
NB = W / 2tan∅ -------let this be equation 1
∑Fx = 0
⇒ FA - NB = 0
FA = NB
therefore from equation 1
FA = NB = W / 2tan∅
we substitute in our values
FA = NB = 76 / 2tan(60°) = 21.9393 lb
Now ∑Fy = 0
NA - W = 0
NA = W = 76 lb
Net force at A will be
FA' = √( NA² + FA²)
= √( (W)² + (W / 2tan∅)²)
we substitute in our values
FA' = √( (76)² + (21.9393)²)
= √( 5776 + 481.3328)
= √ 6257.3328
FA' = 79.1033 lb
Therefore the magnitude of force at point A is 79.1033 lb
Now maximum possible frictional force at A
FA_max = μ × NA
so, FA_max = 0.4 × 76
FA_max = 30.4 lb
So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.
Therefore since FA < FA_max; Ladder WILL NOT slip
Answer
given,
firm is producing = 2,475 units
output by hiring 50 workers W = $20 per hour
25 units of capital R = $10 per hour
marginal product of labor = 40
marginal product of capital = 25
Firm is not minimizing the cost because the firm use more capital and less labor.
I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance: