A block of mass 8.60 kg is pulled along a horizontal frictionless surface by a cord that exerts a force of 33 N at an angle 29.40 degrees from the horizontal. What is the magnitude of the normal force exerted on the mass by the floor?
I solved for the acceleration, which was 3.34 m/s/s (part one of the question) and I tried solving for Fnormal using Fnorm = mg + Fsintheta but the answer it gave me was incorrect
Fn = Mg-F*sin29.4.
To determine the magnitude of the normal force exerted on the mass by the floor, you need to consider the forces acting on the block.
First, let's break down the given force into horizontal and vertical components.
The horizontal component of the force is F_horizontal = F * cos(theta), where F is the given force and theta is the angle with respect to the horizontal. So, F_horizontal = 33 N * cos(29.40°) = 28.10 N.
The vertical component of the force is F_vertical = F * sin(theta), where F is the given force and theta is the angle with respect to the horizontal. So, F_vertical = 33 N * sin(29.40°) = 15.29 N.
In the absence of any vertical acceleration (since the block is on a horizontal surface), the normal force (F_normal) exerted by the floor must balance the vertical forces acting on the block.
The weight of the block (mg) is vertically downward and can be calculated as m * g, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2). So, mg = 8.60 kg * 9.8 m/s^2 = 84.28 N.
Thus, the equation for the vertical forces is:
F_vertical + mg = F_normal.
Substituting the given values:
15.29 N + 84.28 N = F_normal,
F_normal = 99.57 N.
Therefore, the magnitude of the normal force exerted on the mass by the floor is approximately 99.57 N.
If you obtained an incorrect answer, make sure you properly calculate the vertical component of the force and correctly set up the equation for the vertical forces. Double-check your arithmetic calculations as well.