A block of mass m is pulled along a rough horizontal floor by an applied force as shown. The vertical component of the force exerted on the block by the floor is:
A) mg
B) mg - T cos $\theta$
C) mg + T cos $\theta$
D) mg - T sin $\theta$
E) mg + T sin $\theta$

Question

A block of mass m is pulled along a rough horizontal floor by an applied force  as shown. The vertical component of the force exerted on the block by the floor is: 
A) mg 
B) mg - T cos $\theta$ 
C) mg + T cos $\theta$ 
D) mg - T sin $\theta$ 
E) mg + T sin $\theta$

Quizplus · Accepted Answer

The vertical component of the applied force is zero because the force is applied parallel to the horizontal surface. Therefore, the only vertical forces acting on the block are its weight (mg) and the normal force exerted by the floor. 
The tension force (T) is at an angle θ with the horizontal, so its horizontal component is T cos θ and its vertical component is T sin θ. Since the block is being pulled to the right, the frictional force acting on the block must be to the left and have a magnitude of μN, where N is the normal force. Therefore, the net force acting on the block in the vertical direction is: 
N - mg = 0   (because the block is not accelerating in the vertical direction) 
N = mg   (the normal force is equal in magnitude to the weight of the block) 
The vertical component of the tension force is T sin θ, and it acts upwards. Therefore, the vertical force exerted on the block by the floor is: 
N - T sin θ = mg - T sin θ 
= mg - (11eaafba_af49_abb4_a7a5_9fff97a6b71e_TB3976_11) sin θ 
= mg - T sin F1F1F1S1θF1F1F10

A Block of Mass M Is Pulled Along a Rough θ\thetaθ

A Block of Mass M Is Pulled Along a Rough $\theta$