One end of a uniform 2.5-m-long rod of weight 80 N is supported by a cable at an angle of θ = 33° with the rod. The other end rests against the wall, where it is held by friction as shown in the following figure. (Note: There is no hinge at point A.)
What is the frictional force exerted by the wall on the rod at point A?
summing moments about the wall end: 80(2.5/2)-tension*2.5*sin33=0
solve for tension in that.
now sum moments about the center of the rod: forcefriction*2.5-tension(2.5*sin33)
solve for force friction
To find the frictional force exerted by the wall on the rod at point A, we need to analyze the forces acting on the rod.
Let's break down the forces acting on the rod:
1. Weight (W) of the rod: This is acting vertically downward and has a magnitude of 80 N.
2. Tension force (T) in the cable: This is acting at an angle θ = 33° with the rod.
3. Frictional force (f) exerted by the wall on the rod at point A: This is acting horizontally and opposite to the impending motion.
We can analyze the vertical forces first:
The weight of the rod can be broken down into its vertical and horizontal components as follows:
W_vertical = W * sin(θ)
W_vertical = 80 N * sin(33°)
W_vertical ≈ 80 N * 0.55 ≈ 44 N
Since the rod is in equilibrium, the vertical forces must balance out. Therefore, the vertical component of tension (T_vertical) must be equal to the weight's vertical component (W_vertical).
Now, let's analyze the horizontal forces:
The horizontal component of the tension force is balanced by the frictional force (f):
T_horizontal = f
Next, we need to determine the horizontal component of the tension force (T_horizontal):
The tension force can be broken down into its vertical and horizontal components as follows:
T_horizontal = T * cos(θ)
T_horizontal = T * cos(33°)
Since the horizontal forces are in equilibrium, we have:
T_horizontal = f
Now, we can set up an equation using the above relationships:
T_horizontal = f
T * cos(33°) = f
To find the value of T, we can use the fact that the tension force is balancing out the weight's vertical component:
T_vertical = W_vertical
T * sin(33°) = 44 N
Now, we have a system of equations:
T * cos(33°) = f (Equation 1)
T * sin(33°) = 44 N (Equation 2)
To solve for the frictional force (f), we can divide Equation 1 by cos(33°) and substitute T from Equation 2 into it:
f = T * cos(33°)
f = (44 N / sin(33°)) * cos(33°)
f ≈ 52 N
Therefore, the frictional force exerted by the wall on the rod at point A is approximately 52 N.