A 10 kg block is pressed against a vertical wall and held in place by a force F of 180 N, 37° above the

horizontal. The coefficient of static friction μS=0.20. Find the magnitude and direction of the frictional force.

The answer is 10.3 N downward but I don't know how to approach this.
I know friction force is static coefficient x Normal force. Normal force is perpendicular to the surface. Shouldn't the friction force be 0.2 x 180cos37?

180 sin 37 = 108 N up from force

10 * 9.81 = 98.1 N down from weight

difference is to be made up by friction
108 - 98 = 10 N friction down
now is that less than the friction max allowed?
that is your .2*180 cos 37 = 28.7
so we have much more friction available than we need

To solve this problem, you're correct that the frictional force is given by the formula F_friction = μS * N, where μS is the coefficient of static friction and N is the normal force. However, the normal force is not equal to the force F of 180 N that is applied at a 37° angle.

To determine the normal force, we need to consider the forces acting on the block. There are two vertical forces: the weight of the block (mg) acting downward and the vertical component of the applied force (F * sinθ) acting upward. Since the block is in equilibrium, the sum of these forces must be zero.

So, we can write the equation:

mg + F * sinθ = N

The weight of the block (mg) is given by the formula mass * acceleration due to gravity.
m = 10 kg
g = 9.8 m/s^2

mg = 10 kg * 9.8 m/s^2 = 98 N

Plugging in the values, the equation becomes:

98 N + F * sin37° = N

Simplifying the equation:

F * sin37° = N - 98 N
F * sin37° = N - 98 N
F * sin37° = - 98 N
F = - 98 N / sin37°

Now, we can substitute this value of F into the equation for frictional force:

F_friction = μS * N

F_friction = 0.20 * (-98 N / sin37°)

Calculating this expression, we find:

F_friction ≈ -10.3 N

The negative sign indicates that the frictional force acts in the opposite direction to the applied force, which is downward in this case. Thus, the magnitude of the frictional force is 10.3 N, and its direction is downward.

To approach this problem, we need to first identify the forces acting on the block. In this case, we have the force F applied at an angle of 37° above the horizontal, the weight of the block acting vertically downward, and the frictional force acting parallel to the wall.

To determine the frictional force, we need to consider the maximum static friction. The maximum static friction force can be calculated using the equation:

F_s = μ_s * N

where F_s is the maximum static friction force, μ_s is the coefficient of static friction, and N is the normal force.

The normal force is the force exerted by the wall on the block and is equal in magnitude but opposite in direction to the weight of the block. Therefore, the normal force can be calculated as:

N = mg

where m is the mass of the block and g is the acceleration due to gravity.

In this case, the mass of the block is given as 10 kg, so the weight of the block is:

W = mg = 10 kg * 9.8 m/s^2 = 98 N

The normal force exerted by the wall is equal in magnitude but opposite in direction to the weight of the block:

N = 98 N (upward)

Now we can calculate the maximum static friction force:

F_s = μ_s * N = 0.20 * 98 N = 19.6 N

Since the applied force F is greater than the maximum static friction force F_s (180 N > 19.6 N), the block will start to move. At this point, the frictional force acts opposite to the direction of motion of the block and its magnitude is equal to the force of motion.

The magnitude of the frictional force is 19.6 N since it is equal to the maximum static friction force.

Therefore, the magnitude of the frictional force is 19.6 N.

Now, let's determine the direction of the frictional force. Since the applied force F is directed 37° above the horizontal and the frictional force is opposite to the direction of motion, the frictional force will be directed downward.

Thus, the direction of the frictional force is downward.

Therefore, the magnitude and direction of the frictional force are 19.6 N downward.