Question

2. Find direction and magnitude of kinetic friction in following cases.
F = 10 N
F = 10 N
10 N
37°
1 kg
5 m/s
37°
(ii) 10 m/s +
1 kg
Ms = 0.5
My = 0.4
Us = 0.5
UR = 0.4​

Answer 1

Answer:

F = 10 N

F = 10 N

10 N

37°

1 kg

5 m/s

37°

(ii) 10 m/s +

1 kg

Ms = 0.5

My = 0.4

Us = 0.5

UR = 0.4

Answer 2

Given:

2 Forces F1 and F2 of magnitude 10N each

Velocity of block= 5m/s

Mass of block= 1kg

F1 is inclined at angle 37° (with velocity vector)

F2 is inclined at 180° (with velocity vector)

Coefficient of static friction= 0.5

Coefficient of kinetic friction= 0.4

To find:

Direction and magnitude of kinetic friction

Solution:

Since given that the block is moving, this means that kinetic friction is acting.

F1 is inclined at 37°, its components are 10cos 37° î and 10 sin37° j^

Normal reaction experienced by the block = mg + 10 sin37°

= 10 + 6 = 16N         (taking g=10m/s^2)

Kinetic friction = 0.4 X N

=0.4 X 16

= 6.4N

To find direction:

The direction of frictional force is opposite to the net external force acting on it.

Net external force = (10cos 37 - 10)î

= 8-10 î

= 2N (-î)

So the direction of frictional force = +î (in direction of velocity vector)

Hence, the kinetic frictional force acting on block= 6.4N î