Angle of Repose
Trending Questions
- R√1+μ2
- R(1−1√1+μ2)
- R√1+μ2
- R√1+μ2−1
- 30∘
- 45∘
- 60∘
- 75∘
The friction coefficient between a road and the tyre of a vehicle is 43 . Find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped within 5 m.
Two blocks A and B of mass mA and mB respectively are kept in contact on a frictionless table. The experimenter pushes the block A from behind so that the block B, what is the force exerted by the experimenter on A ?
- 0.001 oC
- 0.0025 oC
- 0.05 oC
- 0.0035 oC
The angle between the resultant contact force and the normal force exerted by a body on the other is called the angle of friction. Show that, if λ be the angle of friction and μ the coefficient fo static friction, λ≤tan−1μ
A uniform ladder of length 10.0 m and mass 16.0 kg is resting against a vertical wall making an angle of 37 ∘ with it. The vertical wall is frictionless but the ground is rough. An electrician weighing 60.0 kg climbs up the ladder. If he stays on the ladder at a point 8.00 m from the lower end, what will be the normal force and the force of friction on the ladder by the ground ? What should be the minimum coefficient of friction for the electrician to work safely ?
Based on dimensions, decide which of the following relations for the displacement of a particle undergoing simple harmonic motion is not correct:
- y = a sin(2πtT)
- y = a sin vt
- y = aT sin(ta)
- y = a√2 [sin(2πtT)−cos(2πtT)]
- 6 N
- 6.4 N
- 0.4 N
- zero
The coefficient of static friction between a block of mass m and an incline is μs = 0.3. What can be the maximum angle θ of the incline with the horizontal so that the block does not slip on the plane? (Also called angle of repose)
θ = cot-1 (0.3)
θ = sin-1 (0.3)
θ = cos-1 (0.3)
θ = tan-1 (0.3)
- decreases upto θ=tan−1(μ) and constant after that
- constant upto θ=tan−1(μ) and decrease after that
- Increases upto θ=tan−1(μ) and constant after that
- increases upto θ=tan−1(μ)and decrease after that
- 1 kg
- 2/3 kg
- 3/4 kg
- 1/3 kg
- 4√3+1
- 1√3+2
- 12+√3
- 1+2√3
- A−q, B−s, C−r, D−p
- A−s, B−p, C−r, D−q
- A−r, B−p, C−p, D−q
- A−p, B−q, C−s, D−r
inertia is pmb218 . Find p.
- 13715MR2
- 15215MR2
- 1715MR2
- 20915MR2
If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficient so that a scooter going at 18 km/hr does not skid ?
- IH>IB>IC>IP
- IC>IP>IB>IH
- IP>IH>IB>IC
- none of these
A small mass slides down an inclined plane of inclination with the horizontal. The coefficient friction is where is the distance through which the mass slides down and ​ is a positive constant. Then the distance covered by the mass before it stops is
- 5^i+5^j
- 5^i+5√3^jN
- none of these
- 5√3^i+5^jN
A block is placed on an inclined plane of inclination θ. The angle of inclination is such that the block slides down the plane at a constant speed. The coefficient of kinetic friction between the block and the inclined plane is equal to
sin θ
cos θ
tan θ
cot θ
half of initial value
two times the initial value
one−third of initial value
four times the initial value
- 712ma2
- 112ma2
- 23ma2
- 56ma2
- 9.8 N
- 9.8√3 N
- 4.9√3 N
- 19.6 N
- 2π
⎷((R−r)1.4g)
- 2π√(R−rg)
- 2π√(rRa)
- 2π√(Rgr)
- 0.6 and 0.6
- 0.4 and 0.3
- 0.6 and 0.5
- 0.5 and 0.6