The coefficient of static friction between a block of wood of mass 29.2-kg and a rough table is 0.51. The table is slowly tilted until the block of wood starts to slide down the surface. At what angle of the table with respect to the horizontal does the wood start to slide? Your answer should be given to the nearest tenth of a degree.
Ww = m*g = 29.2kg * 9.8N/kg = 286.2 N. =
Wt. of the wood.
Fp = 286.2*sinA = Force parallel to incline.
Fv = 286.2 *cosA = Force perpendicular to incline.
286.2*sinA-0.51*286.2*cosA = m*a
286.2*sinA -146*cosA = m*0 = 0.
286.2*sinA = 146*cosA
Divide both sides by cosA:
286.2*(sinA/cosA) = 146
Replace sinA/cosA with tanA:
286.2*tanA = 146
tanA = 0.51013.
A = 27.0o.
To find the angle at which the block of wood starts to slide, we can use the concept of equilibrium. When the block is on the verge of sliding, the force of friction just reaches its maximum value.
First, let's find the maximum force of static friction (F_s max) acting on the block.
F_s max = coefficient of static friction * normal force
The normal force (N) can be found using the weight (W) of the block.
N = W = mass * acceleration due to gravity
N = 29.2 kg * 9.8 m/s^2
N = 285.76 N
Now, we can calculate the maximum force of static friction (F_s max).
F_s max = 0.51 * 285.76 N
F_s max ≈ 145.82 N
When the angle of the table is such that the component of weight (W) parallel to the table exceeds the maximum force of static friction, the block will start to slide.
The component of weight parallel to the table is W_parallel = W * sin(theta), where theta is the angle of the table with respect to the horizontal.
Setting W_parallel equal to F_s max, we can solve for theta.
W_parallel = F_s max
W * sin(theta) = F_s max
Using the weight equation W = mass * acceleration due to gravity, we have:
mass * acceleration due to gravity * sin(theta) = F_s max
29.2 kg * 9.8 m/s^2 * sin(theta) = 145.82 N
Now, we can solve this equation for theta:
sin(theta) = 145.82 N / (29.2 kg * 9.8 m/s^2)
sin(theta) ≈ 0.52
To find the angle theta, we can take the inverse sine (or arcsine) of 0.52:
theta ≈ arcsin(0.52)
theta ≈ 31.7 degrees
Therefore, to the nearest tenth of a degree, the angle at which the wood starts to slide is approximately 31.7 degrees.