A box shown in Fig 4-55 lies on an inclined plane tilted at an angle θ to the horizontal. The coefficient of s?
A box shown in Fig 4-55 lies on an inclined plane tilted at an angle θ to the horizontal. The coefficient of static friction between the box and the inclined plane is 0.33 and the coefficient of kinetic friction between the two surfaces is 0.21. The angle θ is initially zero and gradually increased until the box just begins to slide.
Figure 4-55
(a) Calculate the critical value θ.(The angle at which the box just begins to slide)°
(b) Calculate the acceleration of the box at this angle.m/s2
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(A) This angle is known as angle of repose and is given by arctan(µs)
mg cos Θ = Normal
mg sin Θ = friction = µs N = µ mg cos Θ
=> tan Θ = µs
arctan (1 / 3) = 18.43° approx
(B) We assume that kinetic friction acts at this point
=> ma = mg sin Θ - µ mg cos Θ................................(FBD of box)
=> a = g sin Θ - µ g cos Θ
=> a = 10 (1 / √10) - 0.21 * 10 * (3 / √10).........................(Taking g as 10)
=> a = (√10)(1 - 0.63) = 0.37√10
=> a = 1.17 m/s²
(a) By F(net) = F(gravity) - F(friction)
=>for critical θ
=>F(net) = 0
=>F(gravity) = F(friction)
=>mgsinθ = µs x N
=>mgsinθ = µs x mgcosθ
=>tanθ = 0.33 = tan18.3*
=>θ = 18*3'
(b) F(net) = F(gravity) - F(friction)
=>ma = mgsinθ - µk x N
=>ma = mgsinθ - µk x mgcosθ
=>a = gsinθ - µk x gcosθ
=>a = 9.8 x sin30* - 0.21 x 9.8 x cos30*
=>a = 3.12 m/s^2