1. A block is pulled to the right at constant velocity by a 20N force acting at 30o above the horizontal. If the coefficient of sliding friction is 0.5, what is the weight of the block?

explain briefly?

24.64N

Explanation:

The force inclined at 30° exerted on the block has two components the horizontal and vertical components.

The vertical component force is;

FsinA; where A is angle of inclination.

20sin30° = 10N

The horizontal component is;

20cos30° = 17.32N

Now the normal reaction is a vertical upward force acting opposite to the weight.

Note also that for the body to slide it has to overcome a limiting Frictional force and this force must be 17.32N since it's this horizontal component that makes the body slide.

We can then go on to compute the normal reaction from the coefficient of friction formula given as;

Limiting Frictional Force / Normal reaction

By change of subject formula;

Normal reaction = Limiting frictional force / coefficient of friction

By substituting the equivalent values of limiting Frictional force and coefficient of friction, we have:

17.32N/0.5 = 34.64N

Now since the object didn't fall downwards it means the sum of upward force is the same as the sum of down ward force.

The downward forces are the weight of the object and the vertical component of the applied force when the normal reaction is the upward force.

Expressed mathematically, we have:

34.64 = 10 + weight of object

Hence weight of object = 34.64-10 = 24.64N

44.6 N

Explanation:

Draw a free body diagram of the block. There are four forces on the block:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

and applied force F pulling right 30° above horizontal.

Sum of forces in the y direction:

∑F = ma

N + F sin 30° - mg = 0

N = mg - F sin 30°

Sum of forces in the x direction:

∑F = ma

F cos 30° - Nμ = 0

F cos 30° = Nμ

N = F cos 30° / μ

Substitute:

mg - F sin 30° = F cos 30° / μ

mg = F sin 30° + (F cos 30° / μ)

Plug in values:

mg = 20 N sin 30° + (20 N cos 30° / 0.5)

mg = 44.6 N