A force of 70.0 N is exerted at an angle of 30.0° below the horizontal on a block of mass 8.00 kg that is resting on a table. What is the magnitude of the normal force acting on the block?
The upward component of the 70N force is
70.0N*sinTheta. Add that to the downward component, 8g, and you have the net force downward.
To find the magnitude of the normal force acting on the block, we need to consider the forces in the vertical direction.
The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force is the force exerted by the table to support the weight of the block.
In a stationary situation (where the block is not accelerating vertically), the magnitude of the normal force is equal to the weight of the block.
The weight of an object is given by the formula:
Weight = mass * gravity
where gravity is the acceleration due to gravity, which is approximately 9.8 m/s² on the surface of the Earth.
In this question, the mass of the block is given as 8.00 kg, so the weight of the block is:
Weight = 8.00 kg * 9.8 m/s² = 78.4 N
Therefore, the magnitude of the normal force acting on the block is 78.4 N.