Help please!! y-component of force???

Question

You pull with a force of 260 N on a rope that is attached to a block of mass 24 kg, and the block slides across the floor at a constant speed of 1.6 m/s. The rope makes an angle of 25 degrees with the horizontal.

2.)What is the y-component of the force exerted by the Earth on the block?

3.)What is the y-component of the force exerted by the floor on the block (sometimes called the "normal" force, because it is perpendicular to the floor)?

I can get the force of the rope on the block but cant get 2 and 3...any ideas?

Answer 1

If we assume the floor is horizontal, then the force exerted by the earth (gravity) is entirely in the y direction, and is equal to m*g, where m = mass of the block and g = 9.8 m/sec^2

The force exerted by the floor is the force of gravity minus the upward force due to the force on the rope. If the angle the rope makes with the floor is 25º, the y-component is the "side opposite" in a triangle formed by the rope and the y and x axes. The hypotenuse is the force, so the sin of the angle is

sinø = y-component / rope force

so if F = force on the rope

y=component = F * sin(25º)

The total normal force is then

m*g - F*sin(25º), with upwards taken as the positive direction.

Answer 2

The 'y' component is the vertical component= F.sinθ
,........................"..........................." = 260 x Sin 25
.............................."........................= 260 x .4226
..............................'".......................= 109.876 N
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The normal reaction will be equal to the y-component