As shown in the figure below, a uniform beam is supported by a cable at one end
ID: 1291873 • Letter: A
Question
As shown in the figure below, a uniform beam is supported by a cable at one end and the force of friction at the other end. The cable makes an angle of ? = 30°, the length of the beam is L = 2.00 m, the coefficient of static friction between the wall and the beam is ?s = 0.520, and the weight of the beam is represented by w. Determine the minimum distance x from point A at which an additional weight 2w (twice the weight of the rod) can be hung without causing the rod to slip at point A. x = Incorrect: Your answer is incorrect. Since this is an equilibrium problem, we expect to use the first and second conditions of equilibrium to solve the problem. Before starting the problem, a well-drawn figure with all forces and all forces resolved into components would be a real asset to finding a solution. See if you can use the first condition of equilibrium to obtain an expression for the normal force and then the force of friction. Next see if you can use the first condition of equilibrium to obtain an expression for the tension in the cable. Finally see if you can use the second condition of equilibrium to obtain an expression that will allow you to determine the maximum distance you can hang the weight from the end of the beam supported by the wall. m 9-p-017.gif
As shown in the figure below, a uniform beam is supported by a cable at one end and the force of friction at the other end. The cable makes an angle of
? = 30
As shown in the figure below, a uniform beam is supported by a cable at one end and the force of friction at the other end. The cable makes an angle of ? = 30A hat A degree, the length of the beam is L = 2.00 m, the coefficient of static friction between the wall and the beam is ?s = 0.520, and the weight of the beam is represented by w. Determine the minimum distance x from point A at which an additional weight 2w (twice the weight of the rod) can be hung without causing the rod to slip at point A. x = Incorrect: Your answer is incorrect. Since this is an equilibrium problem, we expect to use the first and second conditions of equilibrium to solve the problem. Before starting the problem, a well-drawn figure with all forces and all forces resolved into components would be a real asset to finding a solution. See if you can use the first condition of equilibrium to obtain an expression for the normal force and then the force of friction. Next see if you can use the first condition of equilibrium to obtain an expression for the tension in the cable. Finally see if you can use the second condition of equilibrium to obtain an expression that will allow you to determine the maximum distance you can hang the weight from the end of the beam supported by the wall. m 9-p-017.gif As shown in the figure below, a uniform beam is supported by a cable at one end and the force of friction at the other end. The cable makes an angle of ? = 30 hat A degree, the length of the beam is L = 2.00 m, the coefficient of static friction between the wall and the beam is ?s = 0.520, and the weight of the beam is represented by w. Determine the minimum distancex from point A at which an additional weight 2w (twice the weight of the rod) can be hung without causing the rod to slip at point A. Since this is an equilibrium problem, we expect to use the first and second conditions of equilibrium to solve the problem. Before starting the problem, a well-drawn figure with all forces and all forces resolved into components would be a real asset to finding a solution. See if you can use the first condition of equilibrium to obtain an expression for the normal force and then the force of friction. Next see if you can use the first condition of equilibrium to obtain an expression for the tension in the cable. Finally see if you can use the second condition of equilibrium to obtain an expression that will allow you to determine the maximum distance you can hang the weight from the end of the beam supported by the wall. mExplanation / Answer
along horzantal
N - T*cos30 = 0
N = T*cos30
frictional force f = u*N = u*T*cos30
along vertical
Fnety = 0
f + T*sin30 = W + 2W = 3w
0.52*T*cos30 + T*sin30 = 3w
T = 3w/1.37 = 3.16 w
net torque = 0
w*(L/2) + 2w*x = T*sin30 *L
w*(L/2) + 2w*x = 3.16*w*sin30*L
0.5L + 2x = 1.58 L
x = 0.54 L
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