2 frictin rough inclined plane

FRICTION (rough inclined plane)

Compiled by: RAMAKANT RANA

Q. 1:

Determine the necessary force P acting parallel t

0.25 and pulley to be smooth.

Solution:

Since P is acting downward; the motion too should impend downwards.

Consider first the FBD of 1350N block, as shown in fig (3

∑V = 0

R2 – W = 0

R2 = 1350N

∑H = 0

– T + µR2 = 0

Putting the value of R2 and µ

T = 0.25(1350)

= 337.5N

Now Consider the FBD of 450N block,

∑V = 0

R1 – 450sin45° = 0

R1 = 318.2 N

∑H = 0

T – P + µR1 – 450sin45° = 0

Putting the value of R1, µ and T we get

P = T + µR1 – 450sin45° = 0

= 337.5 + 0.25 × 318.2

P = 98.85 N

Q. 2:

Determine the least value of W in fig(9.38) to keep the

of contact between plane AC and block = 0.28 and that between plane

Solution:


“[email protected]”

acting parallel to the plane as shown in fig 1 to cause motion to impend. µ =

is acting downward; the motion too should impend downwards.

50N block, as shown in fig (3)

...(i)

...(ii)

of 450N block, as shown in fig (2)

...(i)

450sin45° = 0

450sin45° = 0

= 337.5 + 0.25 × 318.2 - 450sin45°

.......ANS

in fig(9.38) to keep the system of connected bodies in equilibrium

and block = 0.28 and that between plane BC and block = 0.02

MAIT

motion to impend. µ =

system of connected bodies in equilibrium µ for surface



For least value W, the motion of 2000N block should be impending downward.

From FBD of block 2000N as shown in fig 12.39

∑V = 0

R1 – 2000cos30° = 0

R1 = 1732.06N

∑H = 0

T + µ1R1 – 2000sin30° = 0

T = 2000sin30° – 0.20 × 1732.06,

Now Consider the FBD of WN block, as shown in fig (9.40)

∑V = 0

R2 = Wcos60° = 0

R2 = 0.5W N

∑H = 0

T – µ2R2 – Wsin60° = 0

653.6 = Wsin60° – 0.28 × 0.5

WLEAST = 649.7N

Q.3:

Block A and B connected by a rigid horizontally bar planed at each end are placed on inclined

shown in fig (1). The weight of the block

just start motion of the system.

Solution:

Let Wa be the weight of block A. Consider the free body diagram of

Assume AB be the Axis of reference.

∑V = 0;

Rsin45° – µBRcos45°

On solving, R = 606.09N

∑H = 0;

C – RCos45° – µBRsin45° = 0

Putting the value of R, we get

C = 557.14N

Where C is the reaction imparted by rod.

Consider the free body diagram of block

∑H = 0;

C + µARcos60° – Rcos30° = 0

Putting all the values we get

R = 751.85N

∑V = 0;

µARsin60° + Rsin60° –

On solving, W = 538.7N ...(vi)

Hence weight of block A = 538.7N



For least value W, the motion of 2000N block should be impending downward.

as shown in fig 12.39

...(i)

2000sin30° = 0

0.20 × 1732.06, T = 653.6N ...(ii)

block, as shown in fig (9.40)

...(iii)

sin60° = 0

0.28 × 0.5W

.......ANS

connected by a rigid horizontally bar planed at each end are placed on inclined

1). The weight of the block B is 300N. Find the limiting values of the weight of the block

. Consider the free body diagram of B. As shown in fig 2. And

cos45° – 300 = 0

...(i)

sin45° = 0 ...(ii)

...(iii)

is the reaction imparted by rod.

Consider the free body diagram of block A as shown in fig 3

cos30° = 0 ...(iv)

...(v)

sin60° – W = 0

.......ANS

MAIT

connected by a rigid horizontally bar planed at each end are placed on inclined planes as

the weight of the block A to



Q. 4: What should be the value of the angle shown in fig 1

the plane? The coefficient of friction for the entire surface = 1/3.

Solution:

Consider the equilibrium of block 30N

∑V = 0;

R1 – 30cosθ = 0,

R1 = 30cosθ

∑H = 0;

T – µR1 – 30sinθ = 0,

T = 10cosθ + 30sinθ

Consider the equilibrium of block 90N

∑V = 0;

R2 – R1 – 90cosθ = 0

R2 = 120cosθ

∑H = 0;

90sinθ – µR1 – µR2 = 0,

90sinθ = 10cosθ + 40sin

tanθ = 5/9 i.e.,

θ θ θ θ = 29.050

Q. 5:

A block weighing 200N is in contact with an inclined plane (Inclination = 30º)

under its own weight. Determine the minimum force applied (1) parallel (2) perpendicular

prevent the motion down the plane. What fo

0.25?

Solution:

Consider the FBD of block as shown in fig 2

From the equilibrium condition

Sum of forces perpendicular to plane = 0

R – Wcos30° = 0;

R = Wcos30°

Sum of forces parallel to plane = 0

µR – Wsin30° = 0

Now body will move down only if the value of µ



e of the angle shown in fig 1 so that the motion of the 90N block

the plane? The coefficient of friction for the entire surface = 1/3.

...(i)

...(ii)

...(iii)

= 0,

+ 40sinθ ...(vi)

block weighing 200N is in contact with an inclined plane (Inclination = 30º) [figure 1]. Will the block

under its own weight. Determine the minimum force applied (1) parallel (2) perpendicular to the plane to

prevent the motion down the plane. What force P will be required to just cause the motion up the plane, µ =

of block as shown in fig 2


...(i)

...(ii)

Now body will move down only if the value of µR is less than Wsin30°

MAIT

so that the motion of the 90N block impends down

. Will the block move

to the plane to

the motion up the plane, µ =



Now, µR = 0.25 × W(0.866) = 0.2165

And Wsin30° = 0.5W

Since value of (iv) is less than value of (

(i) When Force acting parallel to plane as shown in fig 3

Frictional force is acting up the plane

∑V = 0;

R = 0.216W

∑H = 0;

P = Wsin30° – µR

P = 0.5 × 200 – 0.216 × 200

P = 56.7N

(ii) When Force acting perpendicular to plane as shown in fig 4

∑H = 0;

Wsin30° – µR = 0

R = 400

∑V = 0;

P + R = Wcos30°

P = 0.866 × 200 – 400

P = –226.79N

(iii) The force P required to just cause the motion up

plane.

Sum of force perpendicular to plane = 0

R = Wcos30

= 172.2N

Sum of force Parallel to plane = 0

P – µR – Wsin30 = 0

P = 0.25 × 172.2 – 200sin30

P = 143.3N



(0.866) = 0.2165W ...(iii)

...(vi)

) is less than value of (iii) So the body will move down.

el to plane as shown in fig 3


...(v)

0.216 × 200

.......ANS

r to plane as shown in fig 4. Frictional force is acting up the plane

...(vii)

400

.......ANS

) The force P required to just cause the motion up the plane as shown in fig 5. Frictional force

Sum of force perpendicular to plane = 0

...(viii)

.......ANS

MAIT


is acting down the



Q. 6: A body of weight 50KN rests in limiting equilibrium on a rough plane, whose slope is 30º. The

raised to a slope of 45º; what force, applied to the body parallel to the inclined plane,

the plane.

Solution:

Consider when the slope of the plane be 30°


µR – Wsin30° = 0


R – Wcos30° = 0

R = Wcos30°

Putting the value of R in equation (i) We get

µ = tan30º = 0.577

Now consider the case when the slope is 45°, Let Force

In this case


P – µR – Wsin45° = 0


R – Wcos45° = 0

R = Wcos45°

Putting the value of R in equation (iv) we get

P = µWcos45° – Wsin45°

P = 15.20KN

Q. 7: Force of 200N is required just to move a certain body up an inclined plane of angle 150, the

parallel to plane. If angle of indication is

Determine the weight of body and coefficient of friction.

Solution:



: A body of weight 50KN rests in limiting equilibrium on a rough plane, whose slope is 30º. The

raised to a slope of 45º; what force, applied to the body parallel to the inclined plane, will support the body on

the slope of the plane be 30°

...(i)


...(ii)

) We get

...(iii)

Now consider the case when the slope is 45°, Let Force P required to support the body.

sin45° = 0 ...(iv)


...(v)

) we get

sin45°

.......ANS

: Force of 200N is required just to move a certain body up an inclined plane of angle 150, the

parallel to plane. If angle of indication is made 20º the effort again required parallel to plane is found 250N.

Determine the weight of body and coefficient of friction.

MAIT

: A body of weight 50KN rests in limiting equilibrium on a rough plane, whose slope is 30º. The plane is

will support the body on

: Force of 200N is required just to move a certain body up an inclined plane of angle 150, the force being

parallel to plane is found 250N.



Case-1

Consider when the slope of the plane be 15º


P – µR – Wsin15° = 0


R = Wcos15°

Putting the value of (ii) in (i), we get

P = µWcos15° + Wsin15° = 0

200 = 0.96µW + 0.25W

Case-2

Consider When the slope of the plane be 20º


P – µR – Wsin20° = 0


R = Wcos20°

Putting the value of (v) in (iv), We get

P = µWcos20° + Wsin20° = 0

250 = 0.939µW + 0.34W

Solved equation (iii) and (vi) we get

W = 623.6N and µ = 0.06

Q. 8: A four wheel drive can as shown in fig (1

at an angle with the horizontal. If the coefficient of friction between the

maximum inclination that can climb?

Solution:

Let the maximum value for inclination is

Let 0.25m distance is the distance between the inclined surface and C.G. Now,


Wsinθ = µ(R1 + R2)


R1 + R2 = Wcosθ

Putting the value of (ii) in (i), We get

Wsinθ = µ(Wcosθ)

Or µ = tanθ

θ = tan–1(0.3)

θ = 16.69º



the slope of the plane be 15º

...(i)


...(ii)

...(iii)

Consider When the slope of the plane be 20º

...(iv)


...(v)

...(vi)

µ = 0.06 .......ANS

drive can as shown in fig (1) has mass of 2000Kg with passengers. The roadway is inclined

at an angle with the horizontal. If the coefficient of friction between the tires and the road is 0.3, what is the

maximum inclination that can climb?

Let the maximum value for inclination is θ for body to remain stationary.

Let 0.25m distance is the distance between the inclined surface and C.G. Now,

...(i)


...(ii)

.......ANS

MAIT

roadway is inclined

and the road is 0.3, what is the



Q. 9: A weight 500N just starts moving down a rough inclined plane supported by force 200N acting

to the plane and it is at the point of moving up the plane when pulled by a force of

Find the inclination of the plane and the coefficient of friction

Solution:

In first case body is moving down the plane, so frictional force is acting up the plane

Let θ be the angle of inclination and µ be the coefficient of friction.


200 + µR = 500sinθ


R = 500cosθ

Putting the value of (ii) in equation (i)

200 + 500µcos? = 500sin

Now 300N is the force when applied to block, it move in upward direction. Hence in this case f

downward.


R = 500cosθ


300 = µR + 500sinθ

Putting the value of (iv) in equation (v)

300 = 500µcosθ + 500sin

Adding equation (iii) and (vi), We get

Sinθ θ θ θ = 1/2; or θ θ θ θ = 30°

Putting the value in any equation we get

µ = 0.115



moving down a rough inclined plane supported by force 200N acting

to the plane and it is at the point of moving up the plane when pulled by a force of 300N parallel to the plane.

Find the inclination of the plane and the coefficient of friction between the inclined plane and the weight.

In first case body is moving down the plane, so frictional force is acting up the plane

be the angle of inclination and µ be the coefficient of friction.

...(i)


...(ii)

200 + 500µcos? = 500sinθ ...(iii)

Now 300N is the force when applied to block, it move in upward direction. Hence in this case frictional


...(iv)

...(v)

+ 500sinθ ...(vi)

= 30° ........ANS

Putting the value in any equation we get

.......ANS

MAIT

moving down a rough inclined plane supported by force 200N acting parallel

300N parallel to the plane.

tween the inclined plane and the weight.

rictional force acts

2 frictin rough inclined plane

Documents