Question

A 70 kg box on a horizontal surface, is pulled by a force of 400 N at 30 degree with the respect to the horizontal. The coefficient of friction between the box and the surface is 0.50. Assume the motion is horizontal. (a) Draw a free body diagram of the system (b) find the normal force (c) find the horizontal force.

Answer 1

(a) See image attached

(b) 486.7 N

(c) 103.06 N

Step-by-step explanation:

(a)

The free body diagram is attached

(b)

Normal force (N)

`N= mg-Fsin\theta`

Where m is the mass, g for acceleration due to gravity, F for pulling force and
`\theta` is angle of inclination

Substituting 70 Kg for m,
`9.81 m/s^(2)` for g, 400 N for F and
`30^(\circ)` for
`\theta` we obtain

N=(70*9.81)-(400sin 30)=486.7 N

N=486.7 N

(c)

`F_(net)=Fcos\theta-Fr` where
`F_(net)` is the net force on horizontal axis, Fr is frictional force which is given by

`Fr=\mu N`

`F_(net)=Fcos\theta- \mu N`

Substituting 400 for F,
`30^(\circ)` for
`\theta`, 0.5 for
`\mu` and 486.7 for N we obtain

`F_(net)=(400cos30^(\circ))-(0.5*486.7)=103.0601615\approx 103.06 N`

Therefore, horizontal force is 103.06 N