1. A piece of rectangular block weighing 5 kg is pressed on a vertical wall by a horizontal force applied to the block. If the coefficient of friction between the block and the wall is 2.0, what force is necessary to keep the block in place?

2. If the force in Prob. 1 is applied to the block such that it makes 60° with wall, what force is needed to keep the block in place?

down force = m g

up force = 2 F
2 F = 5 (9.81)
F = 5(9.81)/2

Part 2, 60 up or 60 down?

if 60 up

down force = m g

up force = F cos 60 + 2 F sin 60
= .5 F + 1.73 F = 2.23 F
so

2.23 F = 5 * 9.81

F = 5 (9.81) / 2.23

To find the force necessary to keep the block in place, we need to consider the forces acting on the block.

1. In the first problem, since the block is pressed against a vertical wall, the weight of the block acts vertically downwards with a magnitude of 5 kg * 9.8 m/s^2 (acceleration due to gravity) = 49 N. The force necessary to keep the block in place is the horizontal force required to counteract the friction force between the block and the wall.

Now, let's calculate the friction force. The friction force (Ff) can be determined using the formula:

Ff = coefficient of friction * normal force

The normal force (Fn) is the force exerted by the wall on the block perpendicular to the wall's surface. In this case, it is equal to the weight of the block (49 N).

Ff = 2.0 * 49 N = 98 N

Therefore, a force of 98 N is necessary to keep the block in place.

2. In the second problem, where the force is applied at an angle of 60° with the wall, we need to find the component of the force parallel to the wall and use it to determine the necessary force.

The component of the force parallel to the wall can be calculated using the formula:

Force parallel = Force * cos(theta)

Where theta is the angle between the force and the wall.

Force parallel = 98 N * cos(60°)
= 98 N * 0.5
= 49 N

Therefore, a force of 49 N is needed to keep the block in place when the force is applied at an angle of 60° with the wall.