Question

A) the frictional force F newtonsB)The resultant normal reaction of the surface on the metal block

Answer 1

Given:

The mass of the block is.

`m=10\text{ kg}`

The tension on the rope is,

`T=100\text{ N}`

The angle with the horizontal is,

`\theta=60^(\circ)`

The block is moving with constant speed.

as the block is moving with constant speed, the net force on the block will be zero.

Part (A)

we can write in the horizontal direction the component of the tension will be equal to the frictional force and we write,

`\begin{gathered} T\cos 60^(\circ)=F \\ F=100\cos 60^(\circ) \\ F=50\text{ N} \end{gathered}`

Hence the frictional force is 50 N.

\\

Part(B)

The resultant normal reaction will be,

`\begin{gathered} N=T\sin 60^(\circ)-mg \\ =100sin60^(\circ)-10*9.8 \\ =-11.4\text{ N} \end{gathered}`

hence the resultant normal reaction is -11.4 N.