Question

if a load of 1000kg can just be dragged up an incline at 10 degrees to the horizontal by a force of 5KN applied in the most effective direction,what is the value offthe coefficient offriction?

Answer 1

The coefficient of friction is 0.34

Step-by-step explanation:

It is given that,

Mass of the load, m = 1000 kg

It is dragged up an incline at 10 degrees to the horizontal by a force of 5 KN applied in the most effective direction, F = 5 × 10³ N

We need to find the coefficient of friction between the surface and the load. From the attached figure, the load is dragged up with a force of F. A frictional force f will also act in this scenario.

So,
`F=f+mg\ sin\theta`

Since,
`f=\mu N`

or
`f=\mu mg\ cos\theta`

`F=\mu mg\ cos\theta+mg\ sin\theta`

`F-mg\ sin\theta=\mu mg\ cos\theta`

`5* 10^3\ N-1000\ kg* 9.8\ m/s^2\ sin(10)=\mu mg\ cos\theta`

`\mu=(3298.24)/(1000\ kg* 9.8\ m/s^2* cos(10))`

`\mu=0.34`

So, the coefficient of friction is 0.34. Hence, this is the required solution.