Question

A 40 kg crate is pushed across a floor by a force of 10 N at an angle of 35° above the horizontal. Determine the normal force exerted by the floor on the crate.

Answer 1

386.26 N

Step-by-step explanation:

let N be the normal reaction on the block. F is the force, f = 10 N, which is inclined at 35° with the horizontal.

Resolve the force, F into components.

According to the diagram

As the block in the equilibrium in vertical direction so the sum of the forces along the vertical direction remains zero.

So, N + F Sinθ = mg

N = 40 x 9.8 - 10 x Sin 35°

N = 392 - 5.74

N = 386.26 N

Thus, the normal reaction force is 386.26 N.

Answer 2

Answer:389.26 N

Step-by-step explanation:

Given

mass of crate
`m=40\ kg`

Force applied
`F=10\ N`

inclination of force
`\theta =35^(\circ)`

Usually Normal reaction is equal to the weight of the object but here a force is pulling crate in upward direction so effective weight is less than mg

From diagram Normal reaction is

`N=mg-F\sin \theta`

`N=40* 9.8-10\sin (35)`

`N=392-5.73`

`N=389.26\ N`

Same Normal reaction is exerted by crate to the ground and by ground to the crate.