I will be demonstraiting
Control force 1 - I can draw a properly labeled force diagram showing all forces acting on an object, including total force
Control force 2 - I can determine the support force provided by a surface
Control force 3 - I can use interaction and motion concepts to determine unknown forces, accelerations, etc
Control force 4 - I can determine whether the forces are balanced or unbalanced
Control force 5 - I can determine the frictional force between two surfaces
Control force 6 - I can determine the coefficient of friction between two surfaces
Control force 1 - I can draw a properly labeled force diagram showing all forces acting on an object, including total force
Control force 2 - I can determine the support force provided by a surface
Control force 3 - I can use interaction and motion concepts to determine unknown forces, accelerations, etc
Control force 4 - I can determine whether the forces are balanced or unbalanced
Control force 5 - I can determine the frictional force between two surfaces
Control force 6 - I can determine the coefficient of friction between two surfaces
Question
A 20kg wagon is pulled along the level ground by a rope inclined at 30 degrees above the horizontal. A friction force of 30N opposes the motion. How large is the pulling force if the wagon is moving with (a) constant speed and (b) an acceleration of .40m/s^2?
Answers
The forces in problem A are the normal force (N) which is the force of the table on the block, the weight (W) which is the force of the earth on the block, the friction force (f) which is from the table on the block, and the tension force (t) which is the force of the rope on the block. For problem B, the forces are the same but the frictional force (f) is less than the friction force (f) in problem A because the block is in constant motion and the horizontal f and the force need to cancel each other to = 0 if it is moving at a constant speed. The Normal force is less than the weight because Ty+N=W. The tension is bigger than the force but less than W/N forces because of the formula that showed the force for the rope on question a was 35N. The Weight is 9.8x20=196N. For problem B, the f is the smallest because it says in the problem that the box is accelerating and with more friction, there is less acceleration. The weight (W) is the longest force because the normal force and the Ty have to equal the weight (Ty+N=W). The tension force (Tx) is bigger than the friction force because the box is accelerating. The horizontal forces are balanced in problem A because the box is moving at a constant speed. At a constant spec, the forces cancel each other out. The weight and the normal forces are balance because the box is not moving upward or downward. In problem B, the horizontal forces are unbalanced because the box is accelerating. The weight and normal forces are balanced because the box is not moving up or down. The Normal force in problem A is 178.5 because the Ty is pushing up on the block with 17.5 newtons of force and the weigh= 196N. So, the normal force is the only other vertical force which makes the N=178.5N. In problem B, the normal force is 174.5N. It's because the acceleration that occurs increases the Ty force, which decreases the Normal force. The friction for problem A and B is 30N. However, in problem A, the horizontal forces are balanced out because it's moving at a constant speed. In problem B, the horizontal forces aren't balanced out because it's accelerating. In my EF= mxq equation, the Tx and friction forces were left to find out the Tx force because we already knew the friction force, mass, and acceleration. The friction force is -30 because it is the opposing (-force) in the horizontal direction. The m=20 because it said it was a 20kg wagon. The a=.40 because the acceleration is given to you in the problem. I included the full acceleration given which was .40 because they need you to use the initial acceleration. The answer for the horizontal pulling force is 38N. It tells you that there is 38 newtons (N) of force pulling on the object.
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Work.
A. cos30 = 30/h H=35 Newtons B. Total force = m x a -f + tx = 20kg x .40m/s^2 -30 + tx = 20kg x .40m/x^w tx = 38 H= (38/cos30) = 43 Newtons W = N + ty 196 = N + 17.5 W = 178.5 newtons Sin30 = y/43 43 x sin30 = y y=21.5 196=N+21.5 N=174.5 |